IN  MEMORIAM 
FLOR1AN  CAJORI 


Bonaventura  Cavalieri  (1598-1647)  was  one  of  the  most  influential 
mathematicians  of  his  time.  He  was  chiefly  noted  for  his  invention 
of  the  so-called  "  Principle  of  Indivisibles  "  by  which  he  derived  areas 
and  volumes.  See  pages  134  and  202.  frontis 


SOLID  GEOMETRY 


WITH 


PROBLEMS    AND    APPLICATIONS 


REVISED  EDITION 


BY 
H.    E.    SLAUGHT,    PH.D.,    Sc.D. 

PROFESSOR  OF   MATHEMATICS   IN  THE  UNIVERSITY  OF  CHICAGO 
AND 

N.   J.    LENNES,    PH.D. 

PROFESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY 
OF  MONTANA 


ALLYN    AND    BACON 
Boston  Ntto  gorfe  Chicago 


COPYRIGHT,  1919,  BY 
H.  E.  SLAUGHT  AND  N.  J.  LENNES 


f 


PREFACE 

IN  re-writing  the  Solid  Geometry  the  authors  have  consist- 
ently carried  out  the  distinctive  features  described  in  the 
preface  of  the  Plane  Geometry.  Mention  is  here  made  only 
of  certain  matters  which  are  particularly  emphasized  in  the 
Solid  Geometry. 

Owing  to  the  greater  maturity  of  the  pupils  it  has  been  pos- 
sible to  make  the  logical  structure  of  the  Solid  Geometry  more 
prominent  than  in  the  Plane  Geometry.  The  axioms  are 
stated  and  applied  at  the  precise  points  where  they  are  to  be 
used.  Theorems  are  no  longer  quoted  in  the  proofs  but  are 
only  referred  to  by  paragraph  numbers  ;  while  with  increasing 
frequency  the  student  is  left  to  his  own  devices  in  supplying 
the  reasons  and  even  in  rilling  in  the  logical  steps  of  the  argu- 
ment. For  convenience  of  reference  the  axioms  and  theorems 
of  plane  geometry  which  are  used  in  the  Solid  Geometry  are 
collected  in  the  Introduction. 

In  order  to  put  the  essential  principles  of  solid  geometry, 
together  with  a  reasonable  number  of  applications,  within 
limited  bounds  (156  pages),  certain  topics  have  been  placed  in 
an  Appendix.  This  was  done  in  order  to  provide  a  minimum 
course  in  convenient  form  for  class  use  and  not  because  these 
topics,  Similarity  of  Solids  and  Applications  of  Projection,  are 
regarded  as  of  minor  importance.  In  fact,  some  of  the  exam- 
ples under  these  topics  are  among  the  most  interesting  and 
concrete  in  the  text.  For  example,  see  pages  170-172,  177, 
183-184. 

The  exercises  in  the  main  body  of  the  text  are  carefully 
graded  as  to  difficulty  and  are  not  too  numerous  to  be  easily 
performed.  The  concepts  of  three-dimensional  space  are  made 

iii 


M306191 


IV  PREFACE 


clear  and  vivid  by  many  simple  illustrations  and  questions 
under  the  suggestive  headings  "  Sight  Work."  This  plan  of 
giving  many  and  varied  simple  exercises,  so  effective  in  the 
Plane  Geometry,  is  still  more  valuable  in  the  Solid  Geometry 
where  the  visualizing  of  space  relations  is  difficult  for  many 
pupils. 

The  treatment  of  incommensurables  throughout  the  body  of 
this  text,  both  Plane  and  Solid,  is  believed  to  be  sane  and 
sensible.  In  each  case,  a  frank  assumption  is  made  as  to  the 
existence  of  the  concept  in  question  (length  of  a  curve,  area  of 
a  surface,  volume  of  a  solid)  and  of  its  realization  for  all  prac- 
tical purposes  by  the  approximation  process.  Then,  for  theo- 
retical completeness,  rigorous  proofs  of  these  theorems  are 
given  in  Appendix  III,  where  the  theory  of  limits  is  presented 
in  far  simpler  terminology  than  is  found  in  current  text-books 
and  in  such  a  way  as  to  leave  nothing  to  be  unlearned  or  com- 
promised in  later  mathematical  work. 

Acknowledgment  is  due  to  Professor  David  Eugene  Smith 
for  the  use  of  portraits  from  his  collection  of  portraits  of 
famous  mathematicians. 

H.  E.  SLAUGHT 
N.  J.  LENNES 
CHICAGO  and  MISSOULA, 
May,  1919. 


CONTENTS 

• 

PAGE 

INTRODUCTION 1 

Space  Concepts 1 

Axioms  and  Theorems  from  Plane  Geometry     ....  5 

BOOK  I.     PROPERTIES  OF  THE  PLANE    ......  9 

Perpendicular  Planes  and  Lines 10 

Parallel  Planes  and  Lines 20 

Dihedral  Angles 28 

Constructions  of  Planes  and  Lines      ......  35 

Polyhedral  Angles 40 

BOOK  II.     REGULAR  POLYHEDRONS 49 

Construction  of  Regular  Polyhedrons 62 

BOOK  III.     PRISMS  AND  CYLINDERS 53 

Properties  of  Prisms    .........  54 

Properties  of  Cylinders        . .69 

BOOK  IV.     PYRAMIDS  AND  CONES 79 

Properties  of  Pyramids 80 

Properties  of  Cones 92 

BOOK  V.     THE  SPHERE 105 

Spherical  Angles  and  Triangles  .......  117 

Area  of  the  Sphere 136 

Volume  of  the  Sphere 141 

v 


VI  CONTENTS 


APPENDIX  TO  SOLID  GEOMETRY 

I.   Similar  Solids 157 

II.   Applications  of  Projection 178 

III.   Theory  of  Limits 185 

INDEX 207 

• 

PORTRAITS  AND  BIOGRAPHICAL  SKETCHES 

Cavalieri Frontispiece 

FACING    PAGE 

Thales 48 

Archimedes 104 

Legendre 156 


SOLID   GEOMETRY 


SOLID   GEOMETRY 

INTRODUCTION 

1,  Two-Dimensional  Figures.     In  plane  geometry  each  figure 
is   restricted   so  that  all  of  its  parts  lie  in  the  same  plane. 
Such  figures  are  called  two-dimensional  figures. 

A  figure,  all  parts  of  which  lie  in  one  straight  line,  is  a  one-dimen- 
sional figure,  while  a  point  is  of  zero  dimensions. 

2,  Three-Dimensional   Figures.     A   figure,   not    all   parts   of 
which  lie  in  the  same  plane,  is  a  three-dimensional  figure. 

Thus,  a  figure  consisting  of  a  plane  and  a  line  not  in  the  plane  is  a 
three-dimensional  figure  because  the  whole  figure  does  not  lie  in  one 
plane. 

3,  Solid  Geometry  treats  of  the  properties  of  three-dimen- 
sional figures. 

4,  Representation  of  a  Plane.     While  a  plane  is   endless  in 
extent  in  all   its  directions,  it  is  represented  by  a  parallelo- 
gram, or  some  other  limited  plane  figure. 


A  plane  is  designated  by  a  single  letter  in  it,  by  two  letters 
at  opposite  corners  of  the  parallelogram  representing  it,  or  by 
any  three  letters  in  it  but  not  in  the  same  straight  line. 

Thus,  we  say  the  plane  M,  the  plane  PQ,  or  the  plane  ABC. 

1 


SOLID   GEOMETRY 


5,  Figures  in   Plane  and   Solid  Geometry.     In   describing  a 
figure   in   plane    geometry,   it   is    assumed,   usually  without 
special  mention,  that  all  parts  of  the  figure  lie  in  the  same 
plane,  while  in  solid  geometry  it  is  assumed  that  the  whole 
figure  need  not  lie  in  any  one  plane. 

Thus,  in  plane  geometry  we  have  the  theorem  : 

**  Through  a  fixed  point  on  a  line  one  and  only  one  perpendicular  can 
be  drawn  to  the  line." 

If  all  parts  of  the  figure  are  not  required  to  lie  in  one  plane, 
the  theorem  just  quoted  is  far  from 
true.  As  can  be  seen  from  the  figure, 
an  unlimited  number  of  lines  can  be 
drawn  perpendicular  to  a  line  at  a 
point  in  it. 

Thus,  all  the  spokes  of  a  wheel  may  be 
perpendicular  to  the  axle. 

6,  Loci  in  Plane   and  Solid  Geometry.     In   plane   geometry, 
"the  locus  of  all  points   at  a   given  distance   from  a   given 
point"  is   a   circle,  while   in  solid   geometry  this  locus  is  a 
sphere. 


In  plane  geometry,  "  the  locus  of  all  points  at  a  given  dis- 
tance from  a  given  line "  consists  of  two  lines,  each  parallel 


to  the  given  line  and  at  the  given  distance  from  it,  while  in 
solid  geometry  this  locus  is  a  cylindrical  surface  whose  radius 
is  the  given  distance. 


INTRODUCTION  3 


7,  Parallel  Lines.  Skew  Lines.  In  plane  geometry,  two 
lines  which  do  not  meet  are  parallel,  while  in  solid  geometry, 
two  lines  which  do  not  meet  need  not  be  parallel.  That  is, 
they  may  not  be  in  the  same  plane.  Lines  which  are  not 
parallel  and  do  not  meet  are  called  skew  lines. 

In  solid  geometry,  as  in  plane 
geometry,  the  definition  of  parallel 
lines  implies  that  the  lines  lie  in  the 
same  plane.  That  is,  if  two  lines 
are  parallel,  there  is  always  some 
plane  in  which  both  lie.  Thus,  in 
the  figure,  l\  and  Z2  are  parallel,  as 
are  also  li  and  23,  while  £3  and  14  are 
skew. 


SIGHT  WORK 

Note.     In  exercises  1-4  give  the  required  loci  for  both  plane  and  solid 
geometry.    No  proofs  are  required. 

1.  The  locus  of  all  points  six  inches  distant  from  a  given  point. 

2.  The  locus  of  all  points  ten  inches  distant  from  a  given  point. 

3.  The  locus  of  all  points  at  a  perpendicular  distance  of  four  inches 
from  a  given  straight  line. 

4.  The  locus  of  all  points  at  a  perpendicular  distance  of  nine  inches 
from  a  given  straight  line. 

5.  Find  the  locus  of  all  points  one  foot  from  a  given  plane.     Is  this 
a  problem  in  plane  or  in  solid  geometry  ? 

6.  Find  the  locus  of  all  points  equidistant  from  two  parallel  lines  and 
in  the  same  plane  with  them.     Is  this  a  problem  in  plane  or  in  solid 
geometry  ? 

7.  Find  the  locus  of  all  points  equidistant  from  two  given  parallel 
planes.     Is  this  a  problem  in  plane  or  in  solid  geometry? 

8.  The  side  walls  of  your  schoolroom  meet  each  other  in  four  ver- 
tical lines.     Are  any  two  of  these  parallel  ?     Are  any  three  of  them 
parallel  ?    Do  any  three  of  them  lie  in  the  same  plane  ? 

9.  The  side  walls  of  your  schoolroom  meet  the  floor  and  the  ceiling 
in  straight  lines.     Which  of  these  lines  are  parallel  to  each  other  ?    Do 
any  of  these  lines  lie  in  the  same  plane  ? 


SOLID  GEOMETRY 


8,  Representation  of  Solid  Figures  on  a  Plane  Surface.     To  rep- 
resent a  figure  on  a  plane  surface  when  at  least  part  of  the 
figure  does  not  lie  in  that  surface  requires  special  devices. 

Thus,     in     the     parallelogram 
ABCD  used  to  represent  a  plane, 
the  edges  AB  and  BC  are  made 
heavier  than  the  other  two.     This 
indicates  that  the  lower  and  right- 
hand  sides  are  nearer  the  observer 
than  the  other  edges.     Hence,  the 
plane  represented  does  not  lie  in  the  plane  of  the 
paper,  but  the  lower  part  of  it  stands  out  toward 
the  observer. 

The  figure  ABCD  represents  a  triangular  pyra- 
mid. The  corner  marked  B  is  nearest  the  observer 
and  this  is  indicated  by  the  heavy  lines.  The  tri- 
angle ACD  lies  behind  the  pyramid  and  is  thus  far- 
ther from  the  observer.  The  line  AC  is  dotted  to 
indicate  that  it  is  seen  through  the  figure. 

In  the  closed  box  AG,  the  lines  AD,  DC,  and  DH  lie 
behind  the  figure  and  are  dotted,  while  the  others  are 
in  full  view  and  are  solid.  If  the  box  were  open  at  the 
top,  part  of  the  line  DH  would  be  in  full  view  and 
would  be  represented  by  a  solid  line. 

9,  Representation   of  Lines.      The   following 
plan  for  representing  lines  is  generally  adhered 
to  in  this  book : 

(1)  A  line  of  the  main  figure  which  is  not  obscured  by  any  other  part 
of  the  figure  is  represented  by  a  solid  line. 

(2)  An  auxiliary   line,   which  is  drawn  inci- 
dentally in  making  a  proof    or    constructing  a 
figure,  is  marked  in  long  dashes  if  it  is  in  full  view. 

(3)  Any  line  whatever  which  is  behind  a  part 
of  the  figure  is  marked  in  short  dashes  or  dots,  or 
sometimes  is  not  shown  at  all. 

(4)  Where  a  figure  is  shaded  it  is  usually  regarded  as  opaque  and 
the  lines  behind  it  cannot  be  seen  at  all. 

(5)  In  some  cases  a  shaded  surface  is  regarded  as  translucent  and  the 
lines  behind  it  are  seen  dimly.     Such  lines  are  marked  in  short  dashes. 


H 

y 


INTRODUCTION 


The  following  Axioms  and  Theorems  from  plane  geometry 
are  referred  to  in  the  solid  geometry.  The  special  axioms  of 
solid  geometry  will  be  given  as  they  arise  in  the  text. 

AXIOMS 

10.  Things  equal  to  the  same  things  are  equal  to  each  other. 

11.  If  equals  are  added  to  equals,  the  sums  are  equal. 

12.  If  equals  are  subtracted  from  equals,  the  remainders  are 


13.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

14.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

15.  If  equals  are  added  to  unequals,  the  sums  are  unequal  and 
in  the  same  order. 

16.  If  unequals  are  added  to  unequals,  in  the  same  order,  then 
the  sums  are  unequal  and  in  that  order. 

17.  If  equals  are  subtracted  from  unequals,  the  remainders  are 
unequal  and  in  the  same  order. 

18.  If  unequals  are  subtracted  from  equals,  the  remainders  are 
unequal  and  in  the  opposite  order. 

19.  If  a  is  less  than  b  and  b  less  than  c,  then  a  is  less  than  c. 

20.  If  a  and  b  are  quantities  of  the  same  kind,  then  either 
a  >  b,  or  a  =  b,  or  a  <  b. 

21.  Through  a  point  not  on  a  given  line  only  one  straight  line 
can  be  drawn  parallel  to  that  line. 

22.  A  straight   line-segment  is  the  shortest  distance   between 
two  points. 

23.  Corresponding  parts  of  equal  figures  are  equal. 

THEOREMS 

24.  If  two  lines  intersect,  the  vertical  angles  are  equal. 

25.  Two   triangles  are  equal  if  two   sides  and  the   included 
angle  of  one  are  equal  respectively  to  two  sides  and  the  included 
angle  of  the  other. 


6  SOLID  GEOMETRY 

26,  Two  triangles  are  equal  if  two  angles  and  the  included  side 
of  one  are  equal  respectively  to  two  angles  and  the  included  side 
of  the  other. 

27,  Two   triangles  are  equal  if  three  sides  of  one  are  equal 
respectively  to  three  sides  of  the  other. 

28,  Two  points  each  equidistant  from  the  extremities  of  a  line- 
segment  determine  the  perpendicular  bisector  of  the  segment. 

29,  One  and  only  one  perpendicular  can  be  drawn  to  a  line 
through  a  point  whether  that  point  is  on  the  line  or  not. 

30,  The  sum  of  all  consecutive  angles  about  a  point  in  a  plane 
is  four  right  angles. 

31,  The  sum  of  all  consecutive  angles  about  a  point  and  on  one 
side  of  a  straight  line  is  two  right  angles. 

32,  If  two  adjacent  angles  are  supplementary,  their  exterior 
sides  lie  in  the  same  straight  line. 

33,  If  in  two  triangles  two  sides  of  one  are  equal  respectively 
to  two  sides  of  the  other,  but  the  third  side  of  the  first  is  greater 
than  the  third  side  of  the  second,  then  the  included  angle  of  the 
first  is  greater  than  the  included  angle  of  the  second. 

34,  Two   lines  which  are  perpendicular  to  the  same  line  are 
parallel. 

35,  If  a  line  is  perpendicular  to  one  of  two  parallel  lines,  it  is 
perpendicular  to  the  other  also. 

36,  If  two  given  lines  are  perpendicular  respectively  to  each 
of  two  intersecting  lines,  then  the  given  lines  are  not  parallel. 

37,  In  a  right  triangle  there  are  two  acute  angles. 

38,  From  a  point  in  a  perpendicular  to  a  straight  liney  oblique 
segments  are  drawn  to  the  line.     Then, 

(1)  If  the  distances  cut  off  from  the  foot  of  the  perpendicular 
are  unequal,  the  oblique  segments  are  unequal,  that  one  being  the 
greater  which  cuts  off  the  greater  distance;  and 


INTEODUCTION 


(2)  Conversely,  if  the  oblique  segments  are  unequal,  the  dis- 
tances cut  off  are  unequal,  the  greater  segment  cutting  off  the 
greater  distance. 

39,  Two  angles  whose  sides  are  perpendicular,  each  to  each, 
are  equal  or  supplementary. 

40,  Two  right  triangles  are  equal  if  the  hypotenuse  and  a  side 
of  one  are  equal  respectively  to  the  hypotenuse  and  a  side  of  the 
other. 

41,  Two  right  triangles  are  equal  if  a  side  and  an  acute  angle 
of  one  are  equal  respectively  to  the  corresponding  side  and  acute 
angle  of  the  other. 

42,  Two  right  triangles  are  equal  if  the  hypotenuse  and  an 
acute  angle  of  one  are  equal  respectively  to  the  hypotenuse  and  an 
acute  angle  of  the  other. 

43,  A  quadrilateral  is  a  parallelogram 

(1)  if  both  pairs  of  opposite  sides  are  equal ;  or 

(2)  if  two  opposite  sides  are  equal  and  parallel. 

44,  Opposite  sides  of  a  parallelogram  are  equal. 

45,  Two  parallelograms  are  equal  if  an  angle   and  the  two 
adjacent  sides  of  one  are  equal  respectively  to  an  angle  and  the 
two  adjacent  sides  of  the  other. 

46,  The  segment  connecting  the  middle  points  of  the  two  non- 
parallel  sides  of  a  trapezoid  is  parallel  to  the  bases  and  equal  to 
one  half  their  sum. 

47,  The  locus  of  all  points  equidistant  from  the  extremities  of 
a  line-segment  is  the  perpendicular  bisector  of  the  segment. 

48,  In  the  same  circle  or  in  equal  circles  equal  chords  subtend 
equal  arcs. 

49,  A  line  perpendicular  to  a  radius  at  its  extremity  is  tangent 
to  the  circle. 

50,  If  a  line  is  tangent  to  a  circle,  it  is  perpendicular  to  the 
radius  drawn  to  the  point  of  contact. 


8  SOLID  GEOMETRY 


51,  If  in  a  proportion  the  antecedents  are  equal,  then  the  con- 
sequents are  equal  and  conversely. 

52,  In  a  series  of  equal  ratios  the  sum  of  any  two  or  more 
antecedents  is  to  the  sum  of  the  corresponding  consequents  as  any 
antecedent  is  to  its  consequent. 

53,  If  a  line  cuts  two  sides  of  a  triangle  and  is  parallel  to  the 
third  side,  then  any  two  pairs  of  corresponding  segments  form  a 
proportion. 

54,  If  two  sides  of  a  triangle  are  cut  by  a  line  parallel  to  the 
third  side,  a  triangle  is  formed  which  is  similar   io  the   given 
triangle. 

55,  In  two  similar  triangles  corresponding  altitudes  are  pro- 
portional to  any  two  corresponding  sides. 

56,  Two  triangles  are  similar  if  an  angle  of  one  is  equal  to  an 
angle  of  the  other  and  the  pairs  of  adjacent  sides  are  proportional. 

57,  Two  triangles  are  similar  if  their  pairs  of  corresponding 
sides  are  proportional. 

58,  The  area  of  a  parallelogram  is  equal  to  the  product  of  its 
base  and  altitude. 

59,  Two  parallelograms  have  equal  areas  if  they  have  equal 
bases  and  equal  altitudes. 

60,  The  area  of  a  triangle  is  equal  to  one  half  the  product  of 
its  base  and  altitude. 

61,  If  a  is  a  side  of  a  triangle  and  h  the  altitude  on  it  and  b 
another  side  and  k  the  altitude  on  it,  then  ah  =  bJc. 

62,  The  area  of  a  trapezoid  is  equal  to  one  half  the  product  of 
its  altitude  and  the  sum  of  its  bases. 

63,  The  area  of  a  circle  is  one  half  the  circumference  times  the 
radius,  or  in  symbols : 


BOOK  I 


PROPERTIES  OF  THE  PLANE 

64,  Relations  of  Points,  Lines,  and  Planes.     If  a  line  or  a  plane 
contains  a  point,  the  point  is  said  to  be  on  the  line  or  in  the 
plane  and  the  line  or  plane  is  said  to  pass  through  the  point. 
If  a  plane  contains  a  line,  the  line  is  said  to  be  in  the  plane 
and  the  plane  is  said  to  pass  through  the  line. 

65,  AXIOM  1.     If  two  points  of  a  straight  line  lie  in  a  plane 
then  the  whole  line  lies  in  the  plane. 

Since  a  line  is  endless,  it  follows  from  this  axiom  that  a  plane   is 
endless  in  all  its  directions. 

66,  AXIOM  2.     Through  three  non-collinear  points  one  and 
only  one  plane  can  be  passed. 

67,  AXIOM  3.     Two  distinct   planes    cannot 
meet  in  one  point  only. 

68,  Determination  of  a  Plane.     A  plane  is  said  to  be  deter- 
mined by  certain  elements  (lines  or  points)  if  this  plane  contains 
these  elements  while  no  other  plane  does  contain  them. 

While  two  points  determine  a  straight  line  it 
is  obvious  that  two  points  do  not  determine  a  plane. 
The  figure  shows  three  planes,  L,M,N,  all  passing 
through  the  two  points  A  and  B.  But  only  a  cer- 
tain one  of  these  planes  contains  a  given  point  O 
which  is  not  in  the  line  AB. 

We,  therefore,  say  that  three  non-collinear  points  determine  a  plane, 
while  any  number  of  collinear  points  fail  to  determine  a  plane. 

9 


10  /SOLID   GEOMETRY:    BOOK  I 

LINE  COMMON  TO  TWO  PLANES 

69,   THEOKEM  I.     Two  intersecting  planes  meet  in  a 
straight  line. 


/ 


7 


Given  two  intersecting  planes  M  and  N. 

To  prove  that  they  meet  in  a  straight  line  AB. 

Proof:  If  two  planes  intersect  they  meet  in  at  least  two 
points,  as  A  and  B.  Ax.  3,  §  67 

But  A  and  B  determine  a  line  which  lies  wholly  in  M  and 
also  wholly  in  N.  Ax.  1,  §  65 

Hence  the  planes  have  the  straight  line  AB  in  common. 

A  point  C  not  in  AB  cannot  lie  in  both  M  and  N,  for  in  that 
case  the  planes  would  have  three  non-collinear  points  in  com- 
mon and  hence  would  coincide.  Ax.  2,  §  66 

Hence  the  planes  M  and  N  meet  in  the  straight  line  AB. 

Q.  E.  D. 

70,  Foot  of  a  Line  Meeting  a  Plane.     The  point  in  which  a 
straight  line  meets  a  plane  is  called  the  foot  of  the  line. 

71,  Line  and  Plane  Perpendicular  to  Each  Other.     A  -line  is 
said  to  be  perpendicular  to  a  plane  if  it  is  perpendicular  to 
every  line  in  the  plane  passing  through  its  foot.     In  this  case 
the  plane  is  also  said  to  be  perpendicular  to  the  line. 

72,  Line  and  Plane  Oblique  to  Each  Other.     A  line  which  meets 
a  plane  and  is  not  perpendicular  to  it  is 

said  to  be  oblique  to  the  plane.    The  plane 
is  also  said  to  be  oblique  to  the  line. 

In  the  figure,  PA  is  perpendicular  to  the  plane 
M  and  QA  is  oblique  to  it. 


PROPERTIES   OF  THE  PLANE 


11 


ELEMENTS  WHICH  DETERMINE  A  PLANE 

73.  THEOREM  II.  A  plane  is  determined  by  (1)  a 
line  and  a  point  not  on  it,  (2)  two  intersecting  lines, 
and  (3)  two  parallel  lines. 


Given  (1)  a  line  /  and  a  point  P  not  on  it ;  (2)  two  intersecting 
lines  /x  and  /2 ;  (3)  two  parallel  lines  /t  and  L. 

To  prove  that  in  each  case  a  plane  is  deter- 
mined. 

Proof  :  (1)  Let  A  and  B  be  two  points  on  I.  Then 
one  and  only  one  plane  M  can  be  passed  through  I  and 
P  because  one  and  only  one  plane  can  be  passed 
through  A,  B,  and  P.  Ax.  2,  §  66 

(2)  Let  A  be  the  intersection  point  of  l\  and  Z2,  and 
B  and  C  any  other  points,  one  on  li  and  the  other  on 
Then  A,  B,  and  C  determine  the  plane  N  in  which  lie 
li  and  Z2.  Axs.  2,  1.  §§66,  65 

(3)  By  definition  li  and  12  lie  in  a  plane  R.    They 
lie  in  only  one  such  plane  since  the  points  A  and  B 
on  li  and  C  on  £2  lie  in  only  one  plane.  Q.  E.  D. 

74,   COROLLARY  1.     Through  a  line  there  is  more  than  one 
plane. 

Suggestion.  Let  M  be  a  plane  through  the  given 
line  Z,  and  C  a  point  not  in  M.  Then  I  and  C  deter- 
mine a  plane  N  through  I  different  from  M. 


75.   COROLLARY  2.     At  a  point  on   a 
there  is  more  than  one  perpendicular  to  the  line. 

Suggestion.    Let  JW  and  Nloe  planes  each  passing  through  the  given 
line  I.     Then  in  each  plane  there  is  a  line  _L  I  at  any  point  A  on  it. 


12  SOLID  GEOMETRY:    BOOK  I 

LINE  PERPENDICULAR  TO  THE  PLANE  OP  TWO  GIVEN  LINES 

76,  THEOREM  III.  If  a  line  is  perpendicular  to 
each  of  two  lines  at  their  point  of  intersection,  it  is 
perpendicular  to  the  plane  of  these  lines. 


Given  a  line  /  perpendicular  to  each  of  the  lines  ^  and  4  at 
the  point  P. 

To  prove  that  I  is  perpendicular  to  the  plane  of  l±  and  12. 
Proof :  Let  M  be  the  plane  of  ^  and  12,  and  let  13  be  any  line 
in  M  through  P.   Draw  a  line  meeting  Il9  12,  and  Z3  in  the  points 
B,  C,  and  D  respectively.     Let  E  and  F  be  points  on  I,  on 
opposite  sides  of  P,  and  such  that  EP  =  FP.     Draw  EB,  ED, 
EC,  FB,  FD,  FO. 
Then  prove : 

(1)  AEBP=AFBP;  (2)  AECP  =  AFCP; 
(3)  AEBC  =  AFBC;  (4)  A  EBD  =  A  FBD ; 
(5)  AEPD  =  AFPD;  (6)  ^EPD  =  /-FPD. 

.'.  EP  is  perpendicular  to  /3.  Why  ? 

But  J3  is  any  line  in  M  through  P. 

I'-.lineZ  -L  plane  M.  §  71 

Q.  B.  D. 

77.  COROLLARY.  If  each  of  two  lines  is  perpendicular  to  a 
third  line  at  the  same  point,  then  the  plane  of  the  two  lines  is 
perpendicular  to  the  third  line. 


PROPERTIES  OF  THE  PLANE 


13 


SIGHT  WORK 

The  diagram  on  this  page  represents  a  three-dimensional  figure  in  the 
shape  of  an  ordinary  rectangular  box.  In  this  figure  the  points  J.,  IT, 
and  5,  for  instance,  do  not  determine  a  plane,  since  they  all  lie  on  the 
same  straight  line,  while  A,  B,  and  C  do  not  lie  in  a  straight  line,  and 
hence  determine  a  plane. 

1.  In  this  figure  pick  out  several  lines  which  lie  in  one  of  the  surfaces 
and  are  not  obscured  by  the  figure. 

2.  Pick  out  several  lines  which  are  obscured  by  the  figure  ;  also  some 
which  lie  within  the  figure. 

3.  Pick  out  four  sets  of  three  points  each  which  do  not  determine 
planes,  and  also  four  sets  which  do  determine  planes. 


4.  Is  the  line  AB  perpendicular  to  the  plane  BCG  ?    Why  ?    Is  AB 
perpendicular  to  the  plane  AEH  ? 

5.  Pick  out  six  planes  in  the  figure,  each  determined  by  parallel  lines. 

6.  Do  the  points  C,  Z,  E  determine  a  plane  ?  the  points  C,  Z,  G  ? 
the  points  B,  F,  Z  ? 

7.  Using  the  schoolroom,  or  a  room  at  home,  locate  planes  corre- 
sponding to  the  planes  AEG,    KLM,   JVOP,  and  EFG,  in  the  above 
figure. 

8.  Point  out  in  some  room  planes  determined  by  points  corresponding 
to  D,  E,  B  ;  D,  F,  B ;  D,  C,  F  •  A,  B,  H  in  the  above  figure. 


14  SOLID  GEOMETEY:    BOOK  I 

PLANE  PERPENDICULAR  TO  A  LINE 

78,   THEOREM  IV.     Through  a  point   there  is   one 
and  only  one  plane  perpendicular  to  a  line. 


FIG.  1.  FIG,  2, 

Given  a  line  /  and  a  point  P. 

To  prove  that  through  P  there  is  one  and  only  one  plane  J_  I. 

Proof :  (1)    When  the  point  P  is  on  the  line  I.     Fig.  1. 

Through  P  draw  lines  PQ  and  PQ'  both  J_Z.  Then  the 
plane  M,  determined  by  PQ  and  PQ',  is  _L  1.  §  77 

To  prove  that  M  is  the  only  plane  through  P  which  is  _L  I, 
suppose  that  a  plane  M'  through  P  is  also  J_  I.  Let  R  be  a 
plane  through  I  meeting  M  and  M'  in  two  lines.  Then  these 
lines  would  both  lie  in  R  and  be  _L  I,  which  is  impossible.  §  29 

(2)    When  the  point  P  is  not  on  the  line  I     Fig.  2. 

Let  PQ  be  a  line  _L  I  and  let  N  be  a  plane  _L  I  at  Q. 

Then  the  plane  N  passes  through  P  and  is  the  plane 
required. 

For  if  P  does  not  lie  in  N,  then  a  plane  R'  determined  by  I  and 
PQ  cuts  N  in  a  line  V  which  is  JL I  (§  71),  and  PQ  and  V  are 
each  _L  I  at  the  point  Q  in  plane  R',  which  is  impossible.  §  29 

Suppose,  now,  that  there  are  two  planes 
through  P  each  _L  I.  These  planes  cannot 
meet  I  in  the  same  point  (Case  1).  Let  them 
meet  I  in  Q  and  Q'.  Then  PQ  and  PQ'  are 
each  _L  I,  which  is  impossible.  Q.  E.  D. 

79.  COROLLARY.  All  lines  perpendicular  to  a  line  at  the  same 
point  lie  in  the  plane  perpendicular  to  the  line  at  this  point. 


PROPERTIES  OF  THE  PLANE 


15 


LINE  PERPENDICULAR  TO  A  PLANE 

80,   THEOREM  V.     Through  a  point  there  is  one  and 
only  one  line  perpendicular  to  a  plane. 


Given  a  plane  M  and  a  point  P. 

To  prove  that  through  P  there  is  one  and  only  one  line  J_  M. 

Proof  :  (1)  When  P  is  in  the  plane  M  (first  figure) .  Let  li  be  any  line 
in  M  through  P,  and  let  N  be  a  plane  ±  li  at  P  and  meeting  M  in  the  line 
12.  Let  PA  be  a  line  in  N  and  _L  h.  Then  PA  is  also  J_  li.  §  71 

.-.  PJ.JL.ftf.  §76 

(2)  When  P  is  not  in  the  plane  M  (second  figure).  Let  I  be  any  line 
in  the  plane  M .  Through  P  pass  a  plane  N1  ±  I  at  A  and  meeting  M  in 
the  line  AK.  From  P  in  plane  N*  draw  a  line  PO  _L  J.j?T  and  extend  it 
to  P'  so  that  OP'  =  PO.  Let  B  be  any  point  in  I  different  from  A. 
Draw  PB,  PA,  P'B,  P'A,  and  OB. 

Then  prove  (1)  A  POJ.  =  A  P'OA  ;  (2)  A  PAB  =  A  P'AB ;  (3)  A  POB 
=  A  P1  OB  ;  (4)  Z.  POB  =  £  P1  OB  •  (5)  PO  JL  OB. 

.-.PO±M.  §76 

If  in  either  case  (1)  or  case  (2)  there  were  two  lines  PA 
and  PB  each  _L  Jlf,  then  the 
plane  R  of  these  lines  would 
cut  M  in  a  line  I.  Hence 
PA  and  PB  would  both 
lie  in  R  and  be  _L  Z,  which 
also  lies  in  R.  But  this  is 
impossible  by  §  29. 

Hence  PA  is  the  only  line  through  P  which  is  _L  Jf.    Q.  B.  i> 


16 


SOLID   GEOMETRY:    BOOK  I 


SIGHT  WORK 

1.  Does  a  stool  with  three  legs  always  stand  firmly  on  a  flat  floor  ? 
Why? 

2.  Does  a  table  with  four  legs  always  stand  firmly  on  a  flat  floor  ? 
Why  ?     On  what  conditions  will  such  a  table  stand 

firmly  on  a  flat  floor  ? 

3.  If  the  point  C  does  not  lie  in  the  plane  ABD, 
how  many  different  planes   are  determined  by  the 
points  A,  B,   C,  D  ? 

4.  How  many  planes  are  determined  by  any  four 
points  which  do  not  all  lie  in  one  plane  ? 

5.  How  many  planes  are  determined  by  the  points  A,  J3,  C,  D,  E,  if 
A,  JB,  C  lie  in  a  straight  line  and  C,  D,  E  lie  in  another  straight  line  ? 

6.  How  many  planes  are  determined  by  five  points,  no  three  of  which 
lie  in  a  straight  line  and  no  four  of  which  lie  in  the  same  plane  ? 

7.  How  many  planes  are  determined  by  three  lines 
lit  hi  ^3  all  passing  through  the  same  point  but  not  all 
lying  in  the  same  plane  ? 

8.  How  many  planes  are  determined  by  four  lines 
which  all  meet  in  a  point,  but  no  three  of  which  lie  in 
the  same  plane  ? 

9.  How  many  planes  are  determined  by  three  lines 
all  parallel  to  each  other,  and  not  all  lying  in  the  same 
plane  ? 

10.  How  many  planes  are  determined  by  four  lines 

all  parallel  to  each  other,  and  no  three  lying  in  the  same  plane  ? 

11.  A  line  cannot  be  perpendicular  to  each  of  two  intersecting  planes. 
Why? 

Suggestions.  (1)  If  a  line  I  is  perpendicular  to  the  planes  M  and 
the  points  A  and  B,  and  C  is  a  point 
in  their  intersection,  then  A  ABC 
would  contain  two  right  angles. 
(2)  If  I  is  perpendicular  to  M  and  N 
at  a  point  P  in  their  intersection, 
pass  a  plane  through  J,  meeting  M 
and  N  in  h  and  12.  Then  in  this  plane  li  and  12  are  both  _L  I 


PROPERTIES  OF  THE  PLANE  17 

OBLIQUE  LINES  FROM  A  POINT  TO  A  PLANE 

81.  THEOREM  VI.  Oblique  lines  from  a  point  to  a 
plane  meeting  the  plane  at  equal  distances  from  the 
foot  of  the  perpendicular  are  equal ;  and 

conversely,  two  equal  oblique  lines  from  a  point  to 
a  plane  meet  the  plane  at  equal  distances  from  the  foot 
of  the  perpendicular. 


(1)  Given  PC  JL  M,  and  AC  =  BC.     To  prove  that  PA  =  PB. 
Suggestion.     Prove  A  PGA  =  A  PCB. 

(2)  Given  PC  _L  AT,  and  PA  =  PB.     To  prove  that  AC  =  BC. 
Suggestion.     Prove  A  PGA  =  A  PCB. 

82,  COROLLARY.  The  perpendicular  is  the  shortest  distance 
from  a  point  to  a  plane. 

Hence  the  distance  from  a  point  to  a  plane  means  the  per- 
pendicular distance. 

SIGHT  WORK 

Without  giving  proofs  describe  the  following  loci : 

1.  All  points  equidistant  from  the  points  on  a  circle. 

2.  All  points  equidistant  from  the  vertices  of  a  triangle. 

3.  All  points  in  a  plane  which  are  at  a  given  distance  from  a  given 
point  outside  the  plane.     If  a  perpendicular  be  drawn  to  the  plane  from 
this  outside  point,  how  is  its  foot  related  to  this  locus  ? 


18 


SOLID   GEOMETRY:    BOOK  I 


S<*  Base 


i>ase 


EXERCISES 

1.  Show  how  a  carpenter  could  use  the  theorem  of  §  76  to 
stand  a  post  perpendicular  to  the  floor,  if  he  has  at  hand  two 
ordinary  steel  squares. 

2.  Show  how  a  back-stop  on  a  ball  field  can  be  made  per- 
pendicular to  the  line  through 

second  base  and  the  home  plate. 
What  theorems  of  solid  geom- 
etry are  used  ? 

3.  If  a  plane  is  perpendicu- 
lar to  a  line-segment  PP'  at  its 
middle  point,  prove  :  (1)  Every 

point  in  the  plane  is  equally  distant  from  P  and  P' ;  (2)  every 
point  equally  distant  from  P  and  P  lies  in  this  plane.  What 
is  the  locus  of  all  points  in  space  equidistant  from  P  and  P  ? 
Compare  §  47. 

4.  Given  the  points  A  and  B  not  in  a  plane  M.     Find  the 
locus  of  all  points  in  M  equidistant  from  A  and  B. 

Suggestion.     All  such  points  must  lie  in  the  plane  M  and  also  in  the 
plane  which  is  the  perpendicular  bisector  of  the  segment  AB. 

5.  Find  the  locus  of  all  points  equidistant  from  two  given 
points  A  and  B,  and  also  equidistant  from  two  points  C  and  D. 
Discuss.  P 

6.  State   and   prove   a   theorem    of    solid 
geometry  corresponding  to   the   theorem   of 
plane  geometry  given  in  §  38. 

7.  If  in  the  figure   PD  _L  plane   M,  and' 
DC  JL  AB,  a  line  of  the  plane  Jf,  prove  that 
PC±AB. 

Suggestion.     Lay  off  CA  =  CB,  and  compare  tri- 
angles. 

8.  If  in   the    same  figure   PD  J.  M,  and 


PC  J_  AB,  a  line  of  the  plane,  prove  that  DC  JL  AB. 


PROPERTIES   OF  THE  PLANE  19 

PARALLEL  LINES  PERPENDICULAR  TO  THE  SAME  PLANE 

83.    THEOREM  VII.     Two  lines  perpendicular  to  the 
same  plane  are  parallel ;  and 

Conversely,  if  one  of  two  parallel  lines  is  perpen- 
dicular to  a  plane,  the  other  is  also. 

A  c  AC 


FIG.  1.  FIG.  2. 

Given  (1)  AB  and  CD  each  _L  the  plane  M.     Fig.  1. 
To  prove  that  AB  II  CD. 

Proof :  Draw  BD  and  make  DE  _L  DB. 

Take  points  A  and  E  so  that  BA  =  DE,  and  draw  AD,  AE, 
and  BE. 

Now  prove  :  (1)  A  ABD  =  A  BDE  and  .'.  AD  —  BE ; 

(2)  A  ADE  =  A  ABE  and  .-.  Z  ADE  =  Z  ABE  =  Et.  Z. 
.'.  DC,  DA,  DB,  and  BA  all  lie  in  the  same  plane.     §  79 
.'.AB  II  CD.  §34 

Given  (2)  AB  II  CD  and  AB  J_  M.     Fig.  2. 
To  prove  that  CD  _L  3f. 

Proof :  If  CD  is  not  JL  M  let  C"Z>  be  _L  M. 

Then  <7'Z>  II  AB  by  case  (1),  and  C'D  coincides  with  CD  §  21 
and  .'.CD.LM. 

84,  COROLLARY  1.     T/"  eac/i  o/  faco  lines  is 
parallel  to  a  third  line  they  are  parallel  to  each 
other. 

85,  COROLLARY  2.     If  a  plane  is  perpendicu- 
lar to  one  of  two  parallel  lines,  it  is  perpendicu- 
lar to  the  other. 


20  SOLID  GEOMETRY:    BOOK  I 

PARALLEL  PLANES  AND  LINES 

86,  Parallel  Planes.     Two  planes  which  do  not  meet  are  said 
to  be  parallel. 

87,  Line  Parallel  to  a  Plane.     A  straight  line  and  a  plane 
which  do  not  meet  are  said  to  be  parallel. 

88,  Intercepted     Segments.       If    a 
straight  line  Z2  meets  two  planes  in  A 
and  B,  tnen  the  segment  AB  is  said 
to  be  intercepted  by  the  planes. 

Any  line,  as  Zi,  in  either  of  two  parallel  , 
planes,  M  and  JV,  is  parallel  to  the  other  / 
plane.  The  segment  AB  on  the  line  7  'N 


is  intercepted  by  the  planes.  / 

LINE  PARALLEL  TO  A  PLANE 

89,  THEOREM  VIII.  If  a  straight  line  is  parallel 
to  a  given  plane,  it  is  parallel  to  the  intersection  of 
any  plane  through  it  with  the  given  plane. 


N 


IM 


Suggestion  for  proof  .  If  l\  is  the  given  line,  M  the  given  plane,  and 
1%  the  intersection  of  a  plane  JV  through  li  with  M,  show  that  li  and  1%  lie 
in  plane  N  and  cannot  meet. 

90.  COROLLARY  1.     If  a  line  outside  a  plane  is  parallel  to 
some  line  in  the  plane,  then  the  first  line  is  parallel  to  the  plane. 

91.  COROLLARY  2.     If  a  line  is  parallel  to 
a  plane,  then   through  any  point    in  the  plane 
there  is  a  line  in  the  plane  parallel  to  the  given 
line. 

92.  COROLLARY  3.    TTie  intersections  of  a  plane 
with  two  parallel  planes  are  parallel  lines. 


PROPERTIES   OF  THE  PLANE 


21 


PLANES  PERPENDICULAR  TO   A  LINE  ARE  PARALLEL 

93,  THEOREM  IX.  If  each  of  two  planes  is  perpen- 
dicular to  the  same  line,  they  are  parallel ;  and 

Conversely,  if  one  of  two  parallel  planes  is  perpen- 
dicular to  a  line,  the  other  is  also. 


Fia.  1. 


FIG.  2. 


Given  (1)  plane  M  J_  AB  and  plane  N±  AB.     Fig.  1. 

To  prove  that  M  II  N. 

Proof  :  Suppose  M  and  N  to  meet  in  some  point  P.  Draw 
AP  in  M  and  BP  in  N.  Then  AB  _L  AP  and  AB  _L  BP  (§  71), 
which  is  impossible.  §  29 

Given  (2)  M  II  N  and  M±  AB.     Fig.  2. 

To  prove  that  N±.  AB. 

Proof  :  Through  AB  pass  a  plane  cutting  M  and  N  m  AC 
and  5Z>,  and  a  second  plane  cutting  M  and  JV  in  ^4.27  and  BF. 

Then,  ^4C  II  BD  and  AE  II  #F.  §  92 

Now  prove  (1)  AB  J_  ££>,  (2)  AB  _L  JBF.  §  35 

.  §76 


94,  COROLLARY  1.     Parallel   line-segments  included   between 
parallel  planes  are  equal. 

95,  COROLLARY  2.     If  a  line  is  perpendicular  to  one  of  tivo 
parallel  planes,  it  is  perpendicular  to  the  other  also. 

96,  COROLLARY  3.     Two  planes  each  parallel  to  a  third  plane 
are  parallel  to  each  other. 


22  -SOLID   GEOMETEY:    BOOK  I 

PARALLEL   PLANES 

97,  THEOREM  X.  If  a  plane  is  parallel  to  each  of 
two  intersecting  lines,  it  is  parallel  to  the  plane  of  these 
lines. 


M 


Given  a  plane  M  parallel  to  the  intersecting  lines  7j  and  £>. 
To  prove  that  the  plane  M  is  II  the  plane  N  of  l±  and  12- 
Proof :  If  M  is  not  II  N,  these  planes  meet  in  a  line  Z3.      §  69 
Then  neither  l±  nor  12  can  meet  13,  since  they  are  II  M. 
Hence  l±  II  Z3,  and  12  II  13,  which  is  impossible.  §  21 

.•.  M  and  N  cannot  meet  and  are  parallel. 

Q.  E.  D. 

98,   THEOREM  XI.     Through  a  point  not  in  a  plane 
there  is  one  and  only  one  plane  parallel  to  this  plane. 

Given  a  plane  M  and  a  point  P  not  in  M. 

To  prove  that  through  P  there  is  one  and  only 


one  plane  II  M.  IM       \         I 

Proof :  Let  I  be  a  line  through  P  and  J_  M. 
Through  P  draw  lt  and  12  each  J_  1.  §  75 

The  plane  JV  of  ?x  and  Z2  is  J_  I,  and  hence  JV  II  M.     §§  76,  93 
If  through  P  there  were  another  plane  R  II  Jf,  then 
R  would  be  JL  I  at  P.  §  93 

But  N  and  #  cannot  both  be  _L  Z  at  P.  §  78 

Hence  JT  is  the  only  plane  through  P  II  Jf. 

Q.  B.  D. 


PROPERTIES   OF  THE  PLANE  23 

PLANES  PARALLEL  TO  GIVEN  LINES 

99.   THEOREM  XII.     Through  one  of  two  skew  lines 
there  is  one  and  only  one  plane  parallel  to  the  other  line. 


Given  two  skew  lines  /!  and  72.     See  §  7. 
To  prove  that  through  li  there  is  one  and  only  one  plane  II  12. 
Proof :  Through  P,  a  point  in  Il9  draw  a  line  Z3  II  12. 
Then  /x  and  13  determine  a  plane  M  II  12.  §  90 

Any  other  plane  N  through  l±  would  meet  the  plane  of  13  and 
12  in  a  line  through  P  not  II  12  and  hence  N  would  meet  L. 

Q.  E.  D. 

100,  THEOREM  XIII.  Through  a  point  outside  of 
each  of  two  non-parallel  lines  there  is  one  and  only  one 
plane  parallel  to  both  of  these  lines. 


^^-^y 


Given  a  point  P  outside  of  the  non-parallel  lines  /!  and  /2. 

To  prove  that  there  is  one  and  only  one  plane  M  through  P 
parallel  to  li  and  12. 

Proof :    Through  P  pass  Z3  II  ^  and  Z4  II  12. 

Then  the  plane  M  of  13  and  Z4  is  parallel  to  ^  and  12.        §  90 

In  any  other  plane  N  through  P  ||  ^  and  12)  there  are  lines 
1'3  and  l\  through  P  such  that  1'3  li  ^  and  1'4  II  12.  §  91 

In  that  case  Z'3  is  identical  with  13  and  l\  with  Z4.  §  21 

Hence  any  such  plane  ^V  is  identical  with  M.  Q.  E.  r>. 


24  SOLID    GEOMETRY:    BOOK   I 

EXERCISES 

1.  Given  a  plane  M  and  a  point  P  not  in  M.     Find  the  locus 
of  the  middle  points  of  all  segments  connecting  P  with  points 
in  M. 

Suggestion.  Use  the  fact  that  a  line  parallel  to  the  base  of  a  triangle 
and  bisecting  one  side  bisects  the  other  side  also. 

2.  Show  that  a  plane  containing  one  only  of  two  parallel 
lines  is  parallel  to  the  other. 

3.  If  in  two  intersecting  planes  a  line  of  one  is  parallel  to 
a  line  of  the  other,  then  each  of  these  lines  is  parallel  to  the 
line  of  intersection  of  the  planes. 

4.  Show  that  three  lines  which  do  not  meet  in  one  point 
must  all  lie  in  the  same  plane  if  each  intersects  the  other  two. 

5.  Show  that  three   planes,  each   of  which   intersects   the 
other  two,  have  a  point  in  common  unless  their  three  lines  of 
intersection  are  parallel. 

Suggestion.  Suppose  two  of  the  intersection  lines  are  not  parallel,  but 
meet  in  some  point  0.  Then  show  that  the  other  line  of  intersection 
passes  through  O,  and  hence  that  O  is  the  point  common  to  all  three 
planes. 

6.  Given  two  intersecting  planes  M  and  N".     Find  the  locus 
of  all  points  in  M  at  a  given  perpendicular  distance  from  N. 

7.  Given  two  non-intersecting  lines  l±  and  12.     Find  the  locus 
of  all  lines  meeting  Zx  and  parallel  to  12. 

8.  Prove  that  the  middle  points  of  the  sides  of  any  quadri- 
lateral in  space  are  the  vertices  of  a  parallelogram. 

Suggestion.  Use  the  fact  that  a  line  bisecting  two  sides  of  a  triangle 
is  parallel  to  the  third  side.  Note  that  the  four  vertices  of  a  quadrilateral 
in  space  do  not  necessarily  all  lie  in  the  same  plane. 

State  the  corresponding  theorem  in  plane  geometry. 

9.  In  erecting  a  flagpole  on  a  level  space,  show  how  it  can 
be  made  perpendicular  by  means  of  three  ropes  of  equal  length. 
See  §  81. 


PROPERTIES   OF  THE  PLANE  25 

ANGLES   WHOSE   SIDES   ARE  PARALLEL 

101,  THEOREM  XIV.  If  two  intersecting  lines  in 
one  plane  are  parallel,  respectively,  to  two  intersecting 
lines  in  another  plane,  then  the  two  planes  are  parallel, 
and  the  corresponding  angles  formed  ~by  the  lines  are 
equal. 


/         A' 

^^^t  7 

1           A 

c       i 

Given  the  planes  M  and  Nin  which  AB  II  A'B1,  and  AC  II  A'C. 

To  prove  that  M II  N  and  Z  1  =  Z  2. 

Proof :  (1)  If  M  is  not  II  JV,  then  these  planes  meet  in  a 
line  I.  Why? 

Then  neither  AB  nor  AC  can  meet  I  since  each  is  II  N. 

.:  AB  II I  and  AC  II  Z  which  is  impossible.  Why  ? 

(2)  To  prove  Z  1  =  Z  2,  lay  off  AB  =  A'B',  AC  =  A'C',  and 
draw  BC,  B'C',  AA',  BB',  and  CC'. 

Analysis :  Z  1  =  Z  2  if  A  ABC  ==  A  A'B'C',  which  is  true 
if  BC=B'C'. 

But  BC  =  B'C'  if  J^B'C'C  is  a  O,  which  is  so  if  BB'  =  CC' 
and  BB'  II  CC'.  This  last  is  true  if  AA'C'C  and  ^L4'.Br.B  are 
IB,  for  then  ££'  =  AA'  =  CC"  and  J3£'  II  A  A'  II  CC'. 

Hence  we  need  to  prove  in  order  (1)  ^1'C'C  and  AA'B'B 
are  O7,  (2)  ££'C'C  is  a  O,  (3)  A  ABC  =  A  A'B'C',  and 
(4)  Zj&4C  =  Z£^l'C'. 

Hence  it  is  proved  that  the  planes  are  parallel  and  the  angles 
are  equal.  Q.  E.  D. 

102,  COROLLARY.  Two  angles  in  space  whose  sides  are 
parallel  each  to  each  are  either  equal  or  supplementary. 


26  SOLID  GEOMETRY:    BOOK  I 

PARALLEL  PLANES  INTERCEPT  PROPORTIONAL  SEGMENTS 

103,  THEOREM  XV.  If  two  straight  lines  are  cut 
~by  three  parallel  planes,  the  intercepted  segments  on 
one  line  are  proportional  to  the  corresponding  segments 
on  the  other. 


Given  the  lines  AB  and  CD  cut  by  the  planes  AT,  N,  and  P. 


Outline  of  Proof  :  Draw  AD  and  let  the  plane  determined  by 
AD  and  CD  cut  the  planes  M  and  N  in  AC  and  FG  respec- 
tively ;  and  let  the  plane  of  AB  and  AD  cut  N  and  P  in  EF 
and  BD  respectively. 

Then  prove  (1)  FG  II  AC  and  EF  II  BD, 


'  EB     FD          QD     FD'     '  EB      GD 

104,  COROLLARY.  Parallel  planes  which  intercept  equal  seg- 
ments on  any  transversal  line  intercept  equal  segments  on  every 
transversal  line. 

SIGHT  WORK 

Review  the  theorems  of  solid  geometry  up  to  this  point  by  stating  the 
corresponding  theorems  of  plane  geometry,  in  case  such  theorems  exist. 


PROPERTIES   OF  THE  PLANE  27 

EXERCISES 

1.  If  the  spaces  between  four  parallel  shelves  are  5,  8,  and 
10  inches  respectively,  and  a  slanting  rod  intersecting  them 
has  a  7-inch  segment  between  the  first  two  shelves,  find  the 
other  two  segments  of  the  rod. 

2.  If  a  line  cuts  three  parallel  planes,  M,  N,  R,  so  that  the 
segment  intercepted  between  M  and  N  is  7  and  that  between 
N  and  R  is  21,  and  if  another  line  cuts  the  same  planes  so 
that  the  segment  between  M  and  N  is  11,  find  the  segment 
on  the  second  line  between  N  and  R. 

3.  Show    that    line-segments    included    between    parallel 
planes  and  perpendicular  to  them  are  equal,  and  hence  that 
parallel  planes  are  everywhere  equally  distant.     How  can  a 
carpenter  make  use  of  this  principle  in  placing  two  parallel 
shelves  ?     How  many  distances  must  he  measure  ?     Why  ? 

4.  Show  that  through  a  point  outside  a  plane  any  number 
of  lines  can  be  drawn  parallel  to  the  plane.     How  are   all 
these  parallels  related? 

5.  Prove  that  if  a  plane  bisects  two  sides  of  a  triangle  it  is 
parallel  to  the  third  side. 

6.  The  perpendicular  distance  from  a  point  P  to  a  plane 
is  12  in.     Find  the  radius  of  the  circle  which  is  the  locus  of 
all  points  in  the  -plane  at  a  distance  of  20  in.  from  P. 

7.  Show  that,  if  three  line-segments  not  in  the  same  plane 
are  equal  and  parallel,  the  triangles  formed  by  joining  their 
extremities,  as  in  the  figure  of  §  101,  are  equal   and  their 
planes  are  parallel. 

8.  What  is  the  relation  of  two  lines  if  they  are  (a)  parallel 
to  a  given  line,  (b)  perpendicular  to  a  given  line,  (c)  parallel  to 
a  given  plane,  (d)  perpendicular  to  a  given  plane  ? 

9.  What  is   the   relation  of  two  planes  if   they  are  both 
(a)  parallel   to  a   given  plane,   (b)  parallel   to  a  given  line, 
(c)  perpendicular  to  a  given  line  ? 


28 


SOLID   GEOMETRY:    BOOK  I 


DIHEDRAL  ANGLES 

105,  Dihedral  Angle.     The  part  of  a  plane  on  one  side  of  a 
line  in  it  is  called  a  half-plane.     The  line  is  called  the  edge  of 
the  half-plane.     Two  half-planes  meeting  in  a  common  edge 
form  a  dihedral  angle.     The  common  edge  is  the  edge  of  the 
angle  and  the  half-planes  are  its  faces. 

106,  Plane  Angle  of  a  Dihedral  Angle.     Two  lines  in  the  re- 
spective faces  of  a  dihedral   angle  and  perpendicular  to  its 
edge  at  a  common  point  form  a  plane  angle,  which 

is  called  the  plane  angle  of  the  dihedral  angle. 

In  the  figure,  the  half -planes  M  and  N  form  the  dihedral 
angle  M-AB-N,  read  by  naming  the  two  faces  and  the 
edge.  If  CD  in  N  is  _L  AB  and  ED  in  M  is  JL  AB,  then 
Z  CDE  is  the  plane  angle  of  the  dihedral  angle  M-AB-N. 

By  §  101  all  plane  angles  of  a  dihedral  angle 
are  equal  to  each  other. 

107,  Generation  of  a  Dihedral  Angle.     A  dihe- 
dral angle  may  be  thought  of  as  generated  by  the 
rotation   of  a  half-plane   about   its   edge.     The    B 
magnitude  of  the  angle  depends  solely  upon  the 
amount  of  rotation. 

108,  Equal  Dihedral  Angles.     Two  dihedral  angles  are  equal 
when  they  can  be  so  placed  that  they  coincide. 

109,  Right  Dihedral  Angle.     A  right 
dihedral  angle  is  one  whose  plane  angle 
is  a  right  angle. 


110,  Perpendicular  Planes.  Two 
planes  are  said  to  be  mutually  perpen- 
dicular if  their  dihedral  angle  is  a  right  angle. 

Dihedral  angles  are  acute  or  obtuse  according  as  their  plane  angles 
are  acute  or  obtuse. 

Two  dihedral  angles  are  adjacent,  vertical,  supplementary,  or  comple- 
mentary according  as  their  plane  angles  possess  these  properties. 


PROPERTIES  OF  THE  PLANE 


29 


DIHEDRAL  ANGLES  AND  THEIR  PLANE  ANGLES 

111,  THEOKEM  XVI.  Two  dihedral  angles  are 
equal  if  their  plane  angles  are  equal ;  and 

Conversely,  if  two  dihedral  angles  are  equal,  their 
plane  angles  are  equal. 


Given  (1)  the  dihedral  angles  M-AB-N  and  M'-A'B'-N'  in 
which  the  plane  A  CDE  and  C'D'E'  are  equal. 

To  prove  that  Z  M-AB-N =  Z  M'-A'B'-N'. 

Proof :  Place  the  equal  A  CDE  and  C'D'E1  in  coincidence. 

Then  AB  coincides  with  A'B'  since  these  lines  are  perpen- 
dicular to  the  plane  CDE  at  the  point  D.  §  80 

Then  M  and  M '  coincide  as  do  also  N  and  N1  since  they  are 

determined  by  coincident  lines.  §  73 

.-.  Z  M-AB-N =  Z  M'-A'B'-N'.  §  108 

Given  (2)  Z.  M-AB-N -^.M'-A'ff-N'. 

To  prove  that  the  plane  A  CDE  and  C'D'E'  are  equal. 
Proof :  Place  the  equal  dihedral  A  M-AB-N  and  M'-A'B'-N 
in  coincidence  so  that  the  points  D  and  D'  coincide. 

Then  CD  and  C'D'  coincide  as  do  also  ED  and  E'D'.       §  29 
.-.  Z  CDE  =  Z  C'D'E'.  Q.  E.  D. 

112,  COROLLARY.     All  right  dihedral  angles  are  equal. 

113,  Measure  of  a  Dihedral  Angle.     It   follows   from  §  111 
that  the  plane  angle  of  a  dihedral  angle  may  be  regarded  as 
its  measure. 


30 


SOLID   GEOMETRY :    BOOK  I 


MUTUALLY  PERPENDICULAR   PLANES 

114,  THEOREM  XVII.  If  two  planes  are  mutually 
perpendicular,  and  if  a  line  is  drawn  from  a  point  in 
one  perpendicular  to  their  intersection,  then  this  line 
is  perpendicular  to  the  second  plane. 


N 


7 


h/o 


Given  plane  M  _L  plane  N  and  a  point  P  in  M.  Let  /!  be  the 
intersection  of  M  and  N  and  /  a  line  in  M  through  P  J_  /j. 

To  prove  that  I  _L  N. 

Proof:  Let  I  meet  ^  in  O.  Through  O  draw  12  in  JV  and 
perpendicular  to  llt 

Then  the  angle  formed  by  the  lines  I  and  /2  is  the  plane  angle 
of  the  dihedral  angle  between  the  planes  M  and  N.  §  106 


ZJ./2 
l±N. 


§  109 
Why? 

Q.  B.  D. 


115,  COROLLARY.  If  two  planes  are  mutually  perpendicular 
and  if  a  line  is  drawn  from  a  point  in  one  and  perpendicular  to 
the  other,  then  this  line  lies  in  the  first  plane  and  is  perpendicular 
to  the  intersection  of  the  two  planes. 

Proof  :  By  the  theorem  there  is  a  line  through  P  which  lies 
in  the  plane  M  and  is  _L  N.  But  through  P  there  is  only  one 
line  ±.N.  §  80 


PROPERTIES  OF  THE  PLANE 


31 


PLANE  PERPENDICULAR  TO  EACH  OF  TWO  PLANES 

116,  THEOREM  XVIII.  If  a  plane  is  perpendicular 
to  each  of  two  planes,  it  is  perpendicular  to  their  line 
of  intersection. 


Given  Q  JL  M  and  Q  J_  N  and  /  the  intersection  of  M  and  N. 

To  prove  that  Q±.l. 

Proof  :  From  a  point  P  common  to  M  and  N  draw  a  line 


Then  V  lies  in  both  M  and  N. 
Hence,  I'  and  I  are  the  same  line. 
That  is,  I  _L  Q,  or  Q  _L  I 


§  115 
Why  ? 


Q.  B. 


SIGHT  WORK 

1.  State  theorems  on  dihedral  angles  corresponding  to  those  in  §§  24, 
30,  31,  32,  on  plane  angles. 

2.  Name  all  dihedral  angles  in  the  accompanying 
figure. 

3.  Any  plane  _L  the  edge  of  a  dihedral  angle  is 
_L  each  of  its  faces.     Why  ? 

4.  If  each  of  three  lines  is  _L  the  other  two  at 
the  same  point,  then   each  is  _L  the  plane   of  the 
other  two.     Why  ? 

5.  Find  the  locus  of  all  points  at  a  given  distance  from  a  given  plane 
and  also  at  a  given  distance  from  a  second  plane.     Discuss  this  locus  for 
the  various  cases  possible. 


32 


SOLID   GEOMETRY:    BOOK  I 


PERPENDICULAR  LINES   AND  PLANES 

117,  THEOREM  XIX.  If  a  line  is  perpendicular  to 
a  plane,  every  plane  containing  this  line  is  perpendicu- 
lar to  that  plane. 


IN 


Given  a  line  /  J_  plane  N  at  P. 

To  prove  that  a  plane  M  containing  I  is  _L  N. 
Proof :  Let  l±  be  the  intersection  of  M  and  N. 
12  _L  /!  at  P. 

Then  1±11}  l^±lly  andZ_U2. 

.'.MJLN. 


In  N  draw 

Why? 
Why? 


Q.  E.  D. 

118,  Projection  of  a  Figure  on  a  Plane.     The  projection  of  a 
point  on  a  plane  is  the  foot  of  the  perpendicular  from  the 
point  to  the  plane.      The  projection   of 

any  figure  on  a  plane  is  the  locus  of  the 
projections  of  all  points  of  the  figure  on 
the  plane.  / 

Thus,  B  is  the  projection  of  the  point  A  on  M,      I      *• 
and  1%  is  the  projection  of  the  line  li  on  M.  ' 

119,  Angle  Between  Line  and  Plane.     The  angle  between  a 
plane  and  a  line  oblique  to  it  is  understood  to  be  the  acute 
angle  formed  by  the  line  and  its  projection  upon  the  plane. 

120,  Bisector  of  a  Dihedral  Angle.      A  half-plane  bisects  a 
dihedral  angle  if  it  passes  through  the  edge  of  the  dihedral 
angle  and  bisects  its  plane  angle. 


PROPERTIES   OF  THE  PLANE 


PROJECTION   OF   A  STRAIGHT   LINE   ON   A   PLANE 

121,   THEOREM  XX.     The  projection  of  a  straight 
line  on  a  plane  is  a  straight  line  in  that  plane. 


Given  a  plane  M  and  a  line  AB  projected  upon  it. 

To  prove  that  the  projection  is  a  straight  line. 

Proof:  Project  any  three  points  A,  E,  B  of  the  line  AB 
into  the  points  (7,  F,  D  in  the  plane  M. 

Then  AC  and  BD  determine  a  plane  J_  M .  §§  83,  117 

Hence,  by  §§  115,  69,  any  three  points,  C,  F,  D,  of  the  pro- 
jection lie  in  a  straight  line,  that  is,  the  projection  is  a  straight 
line.  Q.  B.  D. 

122,  THEOREM  XXI.  The  acute  angle  formed  ly  a 
straight  line  with  its  own  projection  on  a  plane  is  the 
least  angle  which  it  makes  with  any  line  in  that  plane. 


Given  CB  the  projection  of  AB  on  the  plane  AT,  and  any  other 
line  BD  in  M  through  B. 

To  prove  that  Z  ABD  >  Z  ABC. 

Proof :  Draw  AC  i.  M,  make  BD  =  BC,  and  draw  AD. 

Then  prove  AD  >  AC  and  .'.  Z  ABD  >  Z  ABC.  §  33 


34 


SOLID   GEOMETRY:    BOOK  I 


PLANE  BISECTING  A  DIHEDRAL  ANGLE 

123,  THEOREM  XXII.  The  locus  of  all  points  equally 
distant  from  the  faces  of  a  dihedral  angle  is  the  half- 
plane  bisecting  the  angle. 


Given  the  half-plane  P  bisecting  the  dihedral  Z  M-AB-N. 

To  prove  (1)  that  any  point  E  in  P  is  equally  distant  from  M 
and  N,  and  (2)  that  any  point  which  is  equally  distant  from  M 
and  N  lies  in  P. 

Proof:  (1)     Draw  EC±M  and  ED±N. 

Then  (a)  Plane  CED  J_  M  and  also  J_  N.  Why  ? 

(6)  Plane  CEDJ.AB.  Why? 

Now  Z  COE  is  the  plane  Z  of  M-AB-P, 
and  Z  DOE  is  the  plane  Z  of  N-AB-P.  Why  ? 

.-.  Z  COE  =  Z  DOE  by  the  hypothesis. 

Hence  EC  =  J£D.  §§  42,  23 

(2)  Let  1£  be  any  point  such  that  EC  =  ED,  these  being 
perpendicular  respectively  to  M  and  JV. 

Then  Z  <70#  =  Z  DOE.  §§  40,  23 

Hence,  the  plane  P  determined  by  ABE  is  the  bisector  o 
the  given  dihedral  angle,  that  is,  the  point  E  lies  in  P.     Q.  E.  P 

SIGHT  WORK 

Define  a  locus.  Why  are  two  conditions  necessary  in  determining  a 
locus  ?  Are  these  conditions  fulfilled  in  the  above  theorem  ? 


PROPERTIES  OF  THE  PLANE  35 

PROBLEMS  IN  CONSTRUCTION 

In  making  constructions  in  solid  geometry,  it  is  assumed  that 

124,  A  line  can  be  passed  through  any  two  points. 

125,  A  plane  can  be  passed  through  three  non-collinear  points. 

126,  TJirough  a  point  not  on  a  given  line  there  can  be  drawn 
a  line  parallel  to  this  line. 

127,  Through  a  point  a  line  perpendicular  to  a  given  line  can 
be  drawn. 

128,  COROLLARY.      A  plane  can  be  passed  through  (1)  a  line 
and  a  point  not  on  it,  (2)  two  intersecting  lineSj  (3)  two  parallel 
lines.  §  73 

Q.  E.  F. 

CONSTRUCTING  A  PLANE  PERPENDICULAR  TO  A  LINE 

129,  PROBLEM.     Through  a  given  point  to  construct 
a  plane  perpendicular  to  a  given  line. 


-J 


FIG.  1.  FIG.  2. 

(1)  Given  a  line  /  and  a  point  P  on  it.     Fig.  1. 
To  construct  a  plane  M  _L  I  at  P. 

Construction :  Through  P  draw  two  lines  PA  and  PB  each 

perpendicular  to  J.  §§  127,  75 

Then  the  plane  M  determined  by  PA  and  PB  is  J_  I       §  76 

(2)  Given  a  line  /  and  a  point  P  not  on  it.     Fig.  2. 
To  construct  a  plane  N  through  P  and  J_  I. 

Construction  :  From  P  draw  a  line  PQ  J_  I  §  127 

At  Q  construct  a  plane  NA.I,  as  in  (1). 
Then,  by  §  79,  PQ  lies  in  N,  and  hence  N  is  the  plane  re- 
quired. Q.  B.  F. 


36 


SOLID  GEOMETRY:    BOOK  I 


CONSTRUCTING  A  LINE  PERPENDICULAR  TO  A  PLANE 

130,  PROBLEM.     Through  a  given  point  'to  construct  a 
line  perpendicular  to  a  given  plane. 


K— ~, 


\ 


\. 


(1)  Given  a  plane  M  and  a  point  P  in  it. 
To  construct  a  line  through  P  and  _L  M. 

Construction  :  Let  I  be  any  line  in  M  not  passing  through  P. 
Draw  PB  _L  /.     Through  B  draw  BK  J_  I  but  not  in  M. 
In  the  plane  PBTTdraw  P(7_L  PB  and  meeting  BK  in  C. 
Prove  PC  _L  Jtf.  §§  76,  117,  114 

(2)  Given  a  plane  N  and  a  point  P  not  in  it. 
To  construct  a  line  through  P  and  _L  N. 

Construction  :  Let  I  be  any  line  in  N.     Draw  PB  _L  I  and  in 
JV  draw  BK  J.  Z.     In  the  plane  P£iT  draw  PO_L  £/£ 

Prove,  as  in  (1),  PO±  N.  (See  also  §  80.) 

Q.  E.  F. 

CONSTRUCTING  A  PLANE  PARALLEL  TO  A  LINE 

131,  PROBLEM.     Through  one  of  two  skew  lines  to 
pass  a  plane  parallel  to  the  other. 

Given  two  skew  lines  /!  and  72. 
To  construct  a  plane  through  ^  H  ?2- 
Construction  :  Through  a  point  P  in  ^  pass  a 
line  Z3  1|  12.     (§  126.) 


/  J5/*1    iJ 
/   /"      f 


The  plane  determined  by  ?x  and  lz  is  ||  12 


§  90 

Q.  E.  F. 


PROPEETIES  OF  THE  PLANE 


37 


CONSTRUCTING  A  PLANE  PERPENDICULAR  TO  A  PLANE 

132,  PROBLEM.     Through   a  given   line    to   pass   a 
plane  perpendicular    to   a 

given  plane. 

Given  a  line  /  and  a  plane  M. 

To  pass  a  plane  through  I  and 
JLM. 

Construction  :  Through  a  point  P  in  I  pass  a  line  Z,  _L  M. 

§130 

The  plane  N  determined  by  I  and  ^  passes  through  I  and 
is  JL  M.  §  117 

Q.  E.  F. 

CONSTRUCTING   A  PLANE  PARALLEL   TO  EACH  OF  TWO   LINES 

133,  PROBLEM.     Through  a  point  to  pass   a  plane 
parallel  to  each  of  two  given  lines. 

Given  a  point  P  and  two  lines  l±  and  72. 

To  construct  a  plane  through  P  parallel  to  Zj  and  12. 

Construction :  Through  P  pass  lines  l\  and  Z'2  such  that 
l\  ||  k  and  Z'2 1|  12.  §  126 

Then  the  plane  determined  by  l\  and  1'2  is  parallel  to  both 
Zj  and  12.  §  90 

Q.  E.  F. 

SIGHT  WORK 

1.  Study  the  special  case  of  §  132  in  which  the  given  line  is  perpen- 
dicular to  the  given  plane. 

2.  Is  it  possible  that  the  plane  constructed  by  the  method  of  §  133  may 
contain  one  or  both  of  the  given  lines  ?     Study  all  possible  cases. 

3.  Through  a  given  point  pass  a  plane  _L  each  of  two  given  planes. 
How  many  such  can  be  constructed  ?     Study  the  case  in  which  the  two 
given  planes  are  parallel  to  each  other. 

Suggestion.  Consider  the  relation  of  the  required  plane  to  the  inter- 
section of  the  two  given  planes. 


38  SOLID   GEOMETRY:    BOOK  I 

COMMON  PERPENDICULAR  TO  TWO  SKEW  LINES 

134,  PROBLEM.     To  construct  a  common  perpendicu- 
lar to  two  skew  lines. 


Given  two  skew  lines  l{  and  /2. 

To  construct  a  line  BC  perpendicular  to  each  of  them. 
Construction :  Through  'A,  any  point  in  12,  draw  13  II  /x. 
Let  M  be  the  plane  determined  by  12  and  13. 
Through  Zx  pass  a  plane  N.L  M  and  meeting  M  in  14. 
At  B  the  intersection  of  12  and  14  erect  BC  _L  Jf. 
Then  BC  is  the  required  line  perpendicular  to  ^  and  Z2- 
Outline  of  Proof :  Show  each  of  the  following : 

(1)  BC  lies  in  the  plane  N9  and  .-.  meets  llf 

(2)  li  II Z4,  (3)  BC  J_  Z4,  and  .-.  BC ±  I,. 

135,  COROLLARY.  There  cannot  be  more  than  one  common 
perpendicular  to  two  skew  lines. 

Suggestion.  Suppose  a  second  common  perpendicular  drawn  from  a 
point  PI  in  l\  to  a  point  PZ  in  lz.  Then  show  that  from  PI  it  would  be 
possible  to  have  two  lines  JL  plane  M. 

SIGHT  WORK 

1.  If  a  line  is  _L  a  plane,  show  that  its  projection  is  a  point. 

2.  If  a  line-segment  is  II  a  plane,  show  that  its  projection  is  a  segment 
equal  to  the  given  segment. 

3.  If  a  line-segment  is  oblique  to  a  plane,  show  that  its  projection  is 
less  than  the  given  segment. 


PROPERTIES   OF  THE  PLANE  39 

EXERCISES 

1.  Show  that  the  projections  of  two  parallel  lines  on  a  plane 
are  parallel.     Is  the  converse  true  ?     Illustrate  with  pieces  of 
cardboard. 

2.  If  two  parallel  lines  meet  a  plane,  they  make  equal 
angles  with  it.     Why  ?     Is  the  converse  true  ?     What  is  the 
corresponding  theorem  in  plane  geometry  ?     Is  its  converse 
true? 

3.  If  a  line  cuts  two  parallel  planes,  it  makes  equal  angles 
with  them.     Why  ?     Is  the  converse  true  ?     Discuss  the  cor- 
responding theorems  in  plane  geometry. 

4.  If  two  parallel  line-segments  are  oblique  to  a  plane, 
their  projections  on  the  plane  are  in  the  same  ratio  as  the  given 
segments. 

5.  A  line  and  its  projection  on  a  plane  determine  a  plane 
perpendicular  to  the  given  plane. 

6.  If  a  line  is  parallel  to  one  of  two  given  planes  and  per- 
pendicular to  the  other,  then  the  two  planes  are  perpendicular 
to  each  other. 

7.  Find  the  locus  of  all  points  equidistant  from  two  points 
A  and  B  and  also   at  a  given   distance   from  a  plane  M. 
Discuss. 

8.  Find  the  locus  of  all  points  equidistant  from  two  given 
planes  and  also  equidistant  from  two  given  points.     Discuss. 

9.  Find  the  locus  of  all  points  equidistant  from  two  given 
planes  M  and  N,  and  also  equidistant  from  two  other  given 
planes  Q  and  E.     Discuss.     Compare  with  the  corresponding 
loci  in  plane  geometry. 

10.  Prove  that  there  is  a  line  in  space  every  point  of  which 
is  equidistant  from  three  points  A,  B,  (7,  provided  these  points 
do  not  lie  on  one  line. 


40  -SOLID   GEOMETRY:    BOOK  I 

POLYHEDRAL  ANGLES 

136,  Polyhedral  Angle.     Given  a  convex  polygon  and  a  point 
P  not  in  its  plane.     If  a  half-line  I  with  its  end  point  fixed  at 
P  moves  so  that  it  always  touches  the  polygon  and  is  made  to 
traverse  it  completely,  it  is  said  to  generate  a  convex  polyhedral 
angle. 

137,  Vertex.     Edges.     The  fixed  point  is  the  vertex  of  the 
polyhedral   angle,  and   the   rays   through 

the  vertices  of  the  polygon  are  the  edges 
of  the  polyhedral  angle. 

Any  two  consecutive  edges  determine  a 
plane,  and  the  portion  of  such  a  plane 
included  between  these  edges  is  called  a 
face  of  the  polyhedral  angle. 

138,  Face  Angles.     The  plane  angles  at  the  vertex  are  called 
the  face  angles  of  the  polyhedral  angle.     A  polyhedral  angle 
having  three  faces  is  called  a  trihedral  angle. 

Thus  in  the  figure,  P  is  the  vertex,  PA,  PB,  etc.,  are  the  edges,  and 
/.  APB,  £  BPC,  etc.,  are  the  face  angles. 

139,  Equal  Polyhedral  Angles.     Two  polyhedral  angles  are 
equal  if  they  can  be  made  to  coincide. 

A  polyhedral  angle  is  read  by  naming  the  vertex  and  one  letter  in 
each  edge,  as  P- ABODE,  or  by  naming  the  vertex  alone  where  no  ambi- 
guity would  arise. 

140,  Order  of  Parts  in  Triangles.     In   the  triangles   ABC, 
and  A'B'C',  AB  =  A'B', 

Ttn TIT1'      C* A  r»  A> 

jj\j  —  Jj  w  ,     0^1  —  o  jf±  . 

However,  the  sides  AB, 

BC,    CA,   are   arranged 

in  counter-clockwise  order,  and  the  sides  A  B ,  B  C",  C'A',  are 

arranged  in  clockwise  order.     That  is,  the  parts  of  the  two 

triangles  are  arranged  in  opposite  orders. 


PROPERTIES   OF  THE  PLANE 


41 


141,  Order  of  Parts  in  Polyhedral 
Angles.     In  the  trihedral  angles   0 
and  0'  the  face  angles  AOB,  BOO, 
CO  A  are  arranged  in  counter-clock- 
wise order  as  viewed  from  the  vertex, 
while  A'O'B',  E'O'G',   C'O'A'   are 
arranged  in  clockwise  order. 

The  trihedral  angles  0  and  0'  cannot  be  made  to  coincide,  even  though 
their  corresponding  parts  are  equal.  This  can  be  illustrated  by  attempt- 
ing to  put  a  left  glove  on  the  right  hand. 

CONDITIONS  WHICH  MAKE  TRIHEDRAL  ANGLES  EQUAL 

142,  THEOREM  XXIII.     Two  trihedral  angles   are 
equal   if  two  face  angles  and  the   included  dihedral 
angle  of  one  are  equal  respectively  to  two  face  angles 
and  the  included  dihedral  angle  of  the  other  and  ar- 
ranged in  the  same  order. 


Suggestion  for  Proof  :  This  and  the  following  theorem  may  be 
proved  by  superposition  in  the  same  manner  as  the  correspond- 
ing theorems  on  the  equality  of  triangles  in  plane  geometry. 

143,  THEOREM  XXIV.  Two  trihedral  angles  are 
equal  if  a  face  angle  and  the  tivo  adjacent  dihedral 
angles  of  one  are  equal  respectively  to  a  face  angle  and 
the  adjacent  dihedral  angles  of  the  other  and  arranged 
in  the  same  order. 


42 


SOLID   GEOMETRY:    BOOS  I 


EQUAL  DIHEDRAL  ANGLES  OPPOSITE  EQUAL  FACE  ANGLES 

144,  THEOREM  XXV.  If  two  trihedral  angles  have 
the  three  face  angles  of  the  one  equal  respectively  to  the 
three  face  angles  of  the  other,  the  dihedral  angles  oppo- 
site the  equal  face  angles  are  equal. 

p' 


Given  the  trihedral  angles  P  and  P',  in  which  Z  a  —  Z  a',  Z  b 
c  =  Zc'. 


To  prove  that  the  corresponding  dihedral  angles  are  equal. 

Outline  of  Proof  :  Let  P  and  P'  be  cut  by  the  planes  ABC  and 
AB'C',  making  PA  =  PB=PC  =  P'A'  =  P'B'  ==  FQ'. 

On  the  edges  PA  and  P'A  lay  off  AF=  AF',  and  through 
F  and  F'  pass  planes  _L  to  PA  and  P'A  respectively. 

A  F  and  A'  F1  are  to  be  taken  short  enough  so  that  these  planes  shall 
cut  the  segments  AB,  AC  and  A'B',  A'  C'  respectively. 

In  order  to  prove  that  Z  B-AP-C=^  B'-AP'-C',  i.e.  that 
/.DFE=^D'F'E',  use  the  pairs  of  &APB,  A  P'B'',  BPC, 
B'P'C'  ;  CPA,  C'P'A  ;  ABC,  A'B'C'.     Then  prove  in  order: 
(1)  A  ADF  =  A  AD'F'  ;         (2)  A  AEF  =  A  A'E'F'  ; 
(3)  A  ADE  =  A  A'D'E'  ;         (4)  A  DEF=  A  D'. 
.-.  Z  DFE  =  Z 


145,  COROLLARY.  Two  trihedral  angles  are  equal  if  the  face 
angles  of  one  are  equal  respectively  to  the  face  angles  of  the  other 
and  arranged  in  the  same  order. 


PROPERTIES  OF  THE  PLANE  43 

SYMMETRICAL  TRIHEDRAL  ANGLES 

146,  Symmetrical  Trihedral  Angles.     Two 
trihedral  angles  are  symmetrical,  one  to  the 
other,  if  the  face  angles  and  the  dihedral 
angles  of  one  are  equal  respectively  to  the 
face  angles  and  the  dihedral  angles  of  the 
other,  but  arranged  in  the  opposite  order. 

147,  Vertical  Trihedral  Angles.     Two  tri- 
hedral angles  are  vertical  if  the  edges  of 
one  are  the  prolongations  of  the  edges  of 
the  other. 

148,  COROLLARY  1.     Two  vertical  trihedral  angles  are  sym- 
metrical 

The  proof  is  evident  from  the  figure  above. 

149,  COROLLARY  2.     Two  trihedral  angles  are  symmetrical  if 
the  face  angles  of  one  are  equal  to  the  face  angles  of  the  other  and 
arranged  in  the  opposite  order. 

SIGHT  WORK 

1.   Is  it  possible  that  all  three  face  angles  of  a  trihedral  angle  shall  be 
right  angles  ?    Is  it  possible  that  all  three  dihedral  angles  shall  be  right 


Suggestion.     Consider  a  corner  of  a  rectangular  box. 

2.  If  three  planes  meet  in  a  point  how  many  trihedral  angles  are 
formed  ?    If  all  face  angles  of  one  of  these  trihedral  angles  are  right 
angles,  what  about  the  face  angles  of  the  other  trihedral  angles  ? 

3.  If  in  the  figure  of  §  144  the  Za  =  /.b  =  /.c  =  60°,  what  can  be  said 
about  the  triangles  PAB,  PEG,  PGA,  and  ABC? 

4.  Show  how  to  locate  a  point  which  is  at  a  distance  of  2  feet  from 
each  of  the  three  planes  determined  by  the  faces  of  a  trihedral  angle. 
How  many  such  points  are  there  ? 

Suggestion.    Pass  planes  parallel  to  each  face  of  the  trihedral  angle, 
one  on  each  side  of  it,  and  at  a  distance  of  2  feet  from  it. 


44  -SOLID   GEOMETRY:    BOOK  I 

CONDITIONS  WHICH  MAKE  TRIHEDRAL  ANGLES  SYMMETRICAL 

150,  THEOREM  XXVI.     Two  trihedral  angles   are 
symmetrical 

(1)  if  two  face  angles  and  the   included   dihedral 
angle  of  one  are  equal  respectively  to  two  face  angles 
and  the  included  dihedral  angle  of  the  other,  but  ar- 
ranged in  the  opposite  order  ;  or 

(2)  if  a  face  angle  and  the  adjacent  dihedral  angles 
of  one  are  equal  respectively  to  a  face  angle  and  the 
adjacent  dihedral  angles  of  the  other,  but  arranged  in 
the  opposite  order. 


Proof :  Let  ti  «and  t2  be  two  trihedral  angles  having  the 
properties  specified  under  (1).  Let  t3  be  a  trihedral  angle 
symmetrical  to  ^.  Then  £2  =  ^  §  1^2 

.-.  h  and  tz  are  symmetrical. 

The  proof  of  the  second  part  is  left  to  the  student. 

151,  THEOREM  XXVII.  The  sum  of  two  face  angles 
of  a  trihedral  angle  is  greater  than  the  third  face  angle. 

Outline  of  Proof  :  Connect  points  A  and  C  on  two 
sides.  Suppose  ZADC>Z  ADB.  Construct  DE 
making  Z  ADE  —  Z  ADB.  Suppose  that  point  B  is 
taken  so  that  DB  =  DE,  and  that  AB  and  BG  are 
drawn.  Then  prove 

(1)  AB  =  AE,  (2)  AC<  AB  +  BC, 

(3)  EC  <  BC,  (4)  ^EDC<Z  BDC, 

(5) 


PROPERTIES   OF  THE  PLANE  45 

SUM  OF  THE  FACE  ANGLES  OF   A  POLYHEDRAL  ANGLE 

152,  THEOREM  XXVIII.  The  sum  of  the  face  angles 
of  any  convex  polyhedral  angle  is  less  than  four  right 
angles. 


Proof:  Let  ABODE  be  a  plane  section  of  the  given  poly- 
hedral angle.  The  number  of  triangles  thus  formed  having 
P  for  a  vertex  is  equal  to  the  number  of  face  angles  of  the 
polyhedral  angle. 

Let  0  be  any  point  in  the  base,  and  draw  OA,  OB,  OC,  etc. 

Then  Z  PBA  +  Z  PBQ  >  Z  ABC,  Z.PCB  +  ZPCD> 
Z  BCD,  and  so  on.  §  151 

Now  the  sum  of  the  A  of  the  A  OAB,  OBC,  etc.,  is  equal  to 
the  sum  of  the  A  of  the  A  PAB,  PBC,  etc. 

Hence,  Z  APB  +  ^  BPC  +  •••  <  Z  AOB  +  Z  BOO  +  -. 

But  the  sum  of  the  A  about  0  is  four  right  angles. 

Therefore,  the  sum  of  the  face  angles  of  the  polyhedral 
angle  is  less  than  four  right  angles.  Q.  E.  D. 

Note.  If  as  in  §  144  the  three  edges  of  a  trihedral  angle  are  cut  by  a 
plane,  the  intersection  is  a  triangle.  Then  the  sides  and  angles  of  such 
a  triangle  correspond  to  the  face  angles  and  dihedral  angles  respectively 
of  the  trihedral  angle. 

SIGHT  WORK 

1.  What  theorem  in  plane  geometry  corresponds  to  the  theorem  of 
§  151  in  the  sense  of  the  above  note  ? 

2.  Discuss  propositions  of  plane  geometry  corresponding  to   those 
of  §  150. 


46  SOLID  GEOMETRY:    BOOK  I 

SUMMARY  OF  BOOK  I 

1.  State  the  axioms  used  in  this  Book. 

2.  Describe  the  various  ways  of  determining  a  plane. 

3.  State  the  definitions  on  perpendicular  lines  and  planes. 

4.  State  a  theorem  on  a  line  through  a  point  perpendicular 
to  a  plane. 

5.  State  a  theorem  on  a  plane  through  a  point  perpendicu- 
lar to  a  line. 

6.  State  the  definitions  on  parallel  lines  and  planes. 

7.  State  propositions  on  (1)  a  plane  through  a  point  parallel 
to  a  given  plane ;  (2)  a  plane  through  a  line  parallel  to  a  given 
line ;  (3)  a  plane  through  a  point  parallel  to  two  given  lines. 

8.  State    some   facts   about   perpendiculars   in  the   plane 
which  do  not  hold  in  space. 

9.  Define  a  dihedral  angle  and  the  plane  angle  of  a  dihedral 
angle. 

10.  What  theorems  on  perpendicular  planes  are  proved  in 
connection  with  dihedral  angles  ?     Notice  that  these  theorems 
include  all  those  on  one  plane  perpendicular  to  another.    Why 
should  this  be  the  case  ? 

11.  State  the  definitions  and  theorems  on  projections  given 
in  this  Book. 

12.  State  the  definitions  and  theorems  on  polyhedral  angles 
thus  far  given. 

13.  Define  symmetrical  polyhedral  angles.     Are  symmetrical 
trihedral  angles  equal  ? 

14.  State   theorems   on   equal   and    symmetrical    trihedral 
angles. 

15.  Give  examples  of  loci  in  plane  geometry  which  differ  from 
corresponding  loci  in  solid  geometry.     See,  for  instance,  §  6. 


PROPERTIES   OF  THE  PLANE       •  47 

MISCELLANEOUS  EXERCISES  ON  BOOK  I 

1.  A  Christmas  tree  is  made  to  stand  on  a  cross-shaped 
base.     What  must  be  the  relation  of  the  tree  to  each  piece  of 
the  cross  in  order  that  it  may  be  perpendicular  to  the  floor  ? 

2.  If  A,  B,  and  C  do  not  lie  in  the  same  line,  and  if  their 
projections  on  a  plane  M  do  lie  in  a  straight  line,  what  is  the 
relation  of  the  planes  M  and  ABC? 

3.  Is  it  possible  to  project  a  circle  upon  a  plane  so  that  the 
projection  shall  be  a  straight  line-segment?     If  so,  how  must 
the  circle  and  the  plane  be  related  ? 

4.  If  the  projections  of  a  set  of  points  on  each  of  two  planes 
not  parallel  to  each  other  lie  in  straight  lines,  show  that  the 
points  themselves  lie  in  a  straight  line. 

5.  Find  the  locus  of  points  3  feet  from  one  of  two  inter- 
secting planes  and  6  feet  from  the  other. 

6.  It  is  required  that  a  series  of  electric  lights  shall  be 
7  feet  above  the  floor  of  a  room  and  3  feet  from  the  walls. 
Find  the  locus'  of  all  points  at  which  such  lights  may  be  placed. 

7.  Find  the  locus  of  a  point  in  space  such  that  the  difference 
of  the  squares  of  its  distances  from  two  fixed  points,  A  and  B, 
is  constant. 

Suggestion.    Through  either  A  or  B  construct  a  plane  _L  AB. 

8.  Given   two   points   A  and  B  on  the  same   side   of  a 
plane  M.     Determine  a  point  P  in  M  such  that  AP  -f  PB 
shall  be  a  minimum. 

Suggestion.  Pass  a  plane  through  A  and  B  perpendicular  to  M,  and 
proceed  as  in  Ex.  10,  page  280,  Plane  Geometry. 

9.  Show  that  if  the  edge  of  a  dihedral  angle  is  cut  by  two 
parallel  planes,  the  sections  which  they  make  with  the  faces 
form  equal  angles.     Is  the  converse  proposition  true  ? 

10.  Show  that  if  all  edges  of  a  trihedral  angle  are  cut  by  a 
series  of  parallel  planes,  the  intersections  form  a  series  of 
similar  triangles.  Is  the  converse  proposition  true  ? 


48 


SOLID   GEOMETRY:    BOOK  I 


11.  Find  the  locus  of  the  intersection  points  of  the  medians 
of  the  triangles  obtained  in  Ex.  10.     Also  of  the  altitudes. 

12.  How  many  planes  may  be  made  to  pass  through  a  given 
point  parallel  to  a  given  line  ?     Discuss  the  mutual  relation  of 
all  such  planes. 

13.  Through  a  point  P  construct  a  line  meeting  each  of  two 
lines  Zj  and  12 

Suggestion.     Let  M  and  N  be  the  planes  determined  by  P  and  ii,  P 
and  12.    Consider  their  intersection.   Is  this  construction  always  possible  ? 

14.  A  line  I  is  parallel  to  a  plane  M,  and  lines  ^  and  12  in  M 
are  not  parallel  to  I     Show  that  the  shortest  distance  between 
/  and  Zi  is  equal  to  the  shortest  distance  between  I  and  12. 

15.  Find  the  locus  of  all  points  equidistant  from  the  planes 
determined  by  the  faces  of  a  trihedral  angle. 

16.  Find   the  locus   of  all    points   equi- 
distant from  the  edges  of  a  trihedral  angle. 

Suggestion.  On  the  edges  lay  off  PA=PB=  PC. 
Let  0  be  equidistant  from  J.,  B,  and  C.  Then 
any  point  Q  in  PO  is  equidistant  from  the  edges. 

17.  Planes  determined   by  the  edges  of 
a  trihedral  angle  and  the  bisectors  of  the 
opposite  face  angles  meet  in  a  line. 

18.  If  three  planes  are  so  related  that  the  segments  inter- 
cepted on  any  transversal  line  are  in  the 

same  ratio  as  the  segments  intercepted  on    /       ^ MI 

any  other  transversal  then  the  planes  are 
parallel. 


Suggestion.  Let  Jf,  Nt  Q  be  the  three  planes. 
From  A  any  point  in  M  draw  three  lines,  not  in  the 
same  plane,  meeting  N  in  A*,  Br,  C',  and  Q  in 

A",    B",    C".     Use    the   hypothesis   to    show    that    A'B'  \\A"B"   and 
C'  B'  II  C" B" .    Hence  Q  II  N.     Similarly  show  that  M II  N. 


Thales  of  Miletus  (640-542  B  c.)  was  one  of  the  Seven  Wise  Men  of 
Greece.  He  learned  astronomy  and  geometry  in  Egypt  and  was  the  first 
to  introduce  the  scientific  study  of  geometry  in  Greece.  He  measured 
the  height  of  the  Pyramids  in  Egypt  by  similar  triangles  and  found  a 
method  of  computing  the  distance  of  a  ship  at  sea.  He  predicted  the 
solar  eclipse  of  585  B.C. 


BOOK   II 


REGULAR  POLYHEDRONS 

153,  Polyhedron.     A  polyhedron  is  a  geometric  solid  whose 
boundary  consists  of  plane  polygons. 

154,  Convex    Polyhedron.      A   polyhedron 
is  convex  if  every  section  of  it  made  by  a 
plane  is  a  convex  polygon. 

155,  Faces,  Vertices,  Edges.      The  polygons 
which  bound  the  polyhedron  are  its  faces ; 

the  sides  of  these  faces  are  the  edges  and  their  vertices  are  the 
vertices  of  the  polyhedron. 

156,  Surface  of  Polyhedron.     The  faces,  edges,  and  vertices 
taken  together  form  the  surface  of  the  polyhedron. 

157,  Polyhedrons  Classified  According  to  the  Number  of  Faces. 
A  polyhedron  of  four  faces  is  a  tetrahedron,  one  of  six  faces  is 
a  hexahedron,  one  of  eight  faces  an  octahedron,  one  of  twelve 
faces  a  dodecahedron,  and  one  of  twenty  faces  an  icosahedron. 

These  names  are  all  derived  from  the  Greek  and  refer  to  the  number  of 
faces.  Compare  other  words  derived  from  the  same  Greek  roots  such  as 
hexagon,  octagon,  dodecagon,  etc. 

SIGHT  WORK 

1.  What  is  the  smallest  number  of  polygons  which  may  be  used  to 
inclose  a  portion  of  space  ? 

Suggestion.     Remembering   that  a  triangle  is  re- 
garded as  a  polygon,  consider  a  triangular  pyramid. 

2.  How  many  edges  and  how  many  vertices  has  a 
tetrahedron  ? 

3.  How  many  faces,  edges,  and  vertices  has  a  hexa- 
hedron ? 

Suggestion,     Consider  a  cube. 

49 


50 


SOLID  GEOMETRY. -    BOOK  II 


MODELS  OF  THE  REGULAR  POLYHEDRONS 

158,  Regular  Polyhedrons.  A  polyhedron  whose  polyhedral 
angles  are  equal  and  whose  faces  are  equal  regular  polygons  is 
a  regular  polyhedron.  The  following  are  regular  polyhedrons  : 


TETRAHEDRON 


HEXAHEDRON 


OCTAHEDRON        DODECAHEDRON    ICOSAHEDRON 


159,  Models  of  the  Regular  Polyhedrons.  Models  of  the  regular 
polyhedrons  may  be  made  with  cardboard  by  cutting  out  pat- 
terns, as  shown  in  the  figures  below,  folding  along  the  dotted 
lines,  and  fastening  the  edges  together  by  means  of  gummed 
paper  strips  corresponding  to  the  dotted  margins. 


REGULAR    POLYHEDRONS 


51 


SIGHT  WORK 

1.  What  is  the  smallest  possible  number  of  face  angles  in  a  polyhedral 
angle  ? 

2.  How  many  faces  meet  in  a  vertex  of  a  regular  tetrahedron? 
What  is  the  sum  of  the  face  angles  forming  one  of  its  polyhedral  angles  ? 

C 


3.  How  many  faces  meet  in  a  vertex  of  a  regular  octahedron  ?    in  a 
regular  icosahedron  ? 

4.  What  is  the  sum  of  the  face  angles  forming  a  polyhedral  angle  in 
a  regular  octahedron  ?  in  a  regular  icosahedron  ? 

5.  Why  may  not  six  or  more  equilateral  triangles  be  used  to  form  a 
polyhedral  angle  ? 

6.  What  is  the  sum  of  the  face  angles  forming 
a  polyhedral  angle  of  a  cube  (hexahedron)  ? 

7.  Why  may  not  four  or  more  squares  be  used 
to  form  a  polyhedral  angle  ? 

8.  How  many  faces  meet  in  a  vertex  of  a 
regular  dodecahedron  ?    What  is  the  sum  of  the 
face  angles  forming  one  of  its  polyhedral  angles  ? 

9.  Why  may  not  more  than  three  regular  penta- 
gons be  used  to  form  a  polyhedral  angle  ? 

10.  May  regular  hexagons  be  used  to  form  a  poly- 
hedral angle  ?     Why  ? 

11.  May  any  regular  polygons  of  more  than  six 
sides  be  used  to  form  polyhedral  angles  ?   Why  ? 

12.  Can  we  show  that  not  more  than  three  differ- 
ent   regular    polyhedrons    may    be    formed   having 
triangles  as  their  faces  ? 

13.  How  many  regular  polyhedrons  may  be  formed  having  squares  as 
their  faces  ?    How  many  having  pentagons  as  their  faces  ? 


52 


SOLID   GEOMETRY:    BOOS:  II 


THE  NUMBER  OP  REGULAR  POLYHEDRONS 

160,  THEOREM  I.     There    are    exactly  Jive  regular 
polyhedrons. 

Proof :  On  page  50  it  was  shown  how  models  of  five  different 
regular  polyhedrons  may  be  made. 

Hence  there  are  at  least  five  such  polyhedrons. 

That  there  cannot  be  more  than  five  regular  polyhedrons 
follows  from  the  two  propositions : 

(a)  Every  polyhedral  angle  has  at  least  three  face  angles. 

(b)  The  sum  of  the  face  angles  of  a  polyhedral  angle  is  less  than  360?. 

From  these  propositions  it  follows  that  each  of  the  polyhedral  angles  of 
a  regular  polyhedron  may  be  formed  by  three,  four,  or  five  (but  not  six) 
equilateral  triangles  ;  by  three  (but  not  four)  squares  ;  or  by  three  (but 
not  four)  regular  pentagons.  Regular  polygons  having  more  than  five 
sides  cannot  form  a  polyhedral  angle. 

The  details  of  the  proof  are  left  to  the  student.  The  sight 
work  on  page  51  will  furnish  suggestions. 

161.  Construction  of  Regular  Polyhedrons.     The  regular  poly- 
hedrons may  be  constructed  by  use  of  §  130.     The  construction 
for  the  tetrahedron  and  octahedron  are  given  below. 

(1)  The  regular  tetrahedron.     At  the  center  E  of 
an  equilateral  triangle  ABC  erect  a  perpendicular 
to  the  plane  of  the  triangle.     On  this  take  a  point  D 
so  that  AD  =  AC. 

Then  the  four  triangles,  ABC,  ACD,  ABD,  BCD, 
are  regular  and  equal,  and  the  four  trihedral  angles 
are  equal.  Why  ? 

(2)  The  regular  octahedron.     Through  the  center 
O  of  a  square  ABCD  draw  a  perpendicular  to  the 
plane  of  the  square  and  on  it  take  points  E  and  F 
such  that  AF  =  AE  =  AB.     Join  E  and  F  to  each 

of  the  four  vertices,  A,B,  <7,D. 

Then  the  eight  faces  are  equal  regular  tri- 
angles, and  the  six  polyhedral  angles  are  equal. 
Why? 


BOOK   III 


PRISMS  AND  CYLINDERS 

162,  Closed  Prismatic  Surface.     Given  a  convex  polygon  and 
a  straight  line  not  in  its  plane.     If  the 

line  moves  so  as  to  remain  parallel  to 
itself  while  it  touches  the  boundary  of 
the  polygon  and  is  made  to  traverse  it 
completely,  the  line  is  said  to  generate 
a  closed  prismatic  surface. 

163,  Generator.     Element.     The   mov- 
ing line  is  the  generator  of  the  surface, 
and  the  generator  in  any  one  of  its  posi- 
tions is  an  element  of  the  surface. 

164,  Prism.     A  polyhedron  bounded  by  a  prismatic  surface 
and  two  parallel  plane  sections  cutting  all 

its  elements  is  called  a  prism. 

165,  Bases.      Lateral     Surface.       Edges. 
The    two    parallel    cross-sections    which 
bound  a  prism  are  its  bases  and  the  other 
faces  form  its  lateral  surface.      The  edges 
are  the  lines  in  which  its  lateral  faces  meet. 

166,  Right  Section.     Right  Prism.     A  right  section  of  a  prism 
cuts  all  its  edges  at  right  angles.    A  right  prism 

is  one  whose  bases  are  right  sections. 

167,  COROLLARY  1.     The  lateral  faces  of  a 
prism  are  parallelograms. 

168,  COROLLARY  2.     TJie  lateral  edges  of  a 
prism  are  equal  and  parallel. 


169,   COROLLARY  3.     The  lateral  faces  of  a  right  prism  are 
rectangles. 

53 


54 


SOLID   GEOMETRY:    BOOK  III 


CLASSIFICATION  OP  PRISMS 

170,  Kinds   of  Prisms.     Prisms  are  classified   according   to 
the  form  of  their  right  sections,  as  triangular,  quadrangular, 
pentagonal,  hexagonal,  etc.     A  regular  prism  is  one  whose  right 
section  is  a  regular  polygon. 

171,  Altitude.     Area.     The    altitude    of    a    prism    is    the 
perpendicular  distance  be- 
tween   the    planes    of    its 

bases.  The  altitude  of  a 
right  prism  is  equal  to  its 
edge. 

The  lateral  area  of  a 
prism  is  the  sum  of  the 
areas  of  its  lateral  faces. 

The  total  area  is  the  sum  of  its  lateral  area  and  the  areas 
of  its  bases. 

172,  Truncated    Prism.     A   polyhedron    which 
is  a  part  of  a   prism   cut   off  by  a  plane  meet- 
ing all    the    lateral    edges,   but   not   parallel   to 
the  base,  is  called  a  truncated  prism. 

Two   polyhedrons  are   said  to  be  added  when 
they  are  placed  so  that  a  face  of  one  coincides 
wholly  or  in  part  with  a  face 
of    the  other,    but    otherwise 
each  lies  outside  the  other. 


173,  Parallelepiped.  A  par- 
allelopiped  is  a  prism  whose 
bases,  as  well  as  lateral  faces, 
are  parallelograms. 

A  rectangular  parallelopiped  has  its  bases  and 
all  its  faces  rectangles. 

A   cube  is   a  parallelopiped   whose  bases   and 
faces  are  all  squares. 


PRISMS  AND   CYLINDERS  55 

PARALLEL  CROSS-SECTIONS  OF  A  PRISM  ARE  EQUAL 

174,  THEOREM  I.     The    cross-sections    of   a  prism 
made  by  parallel  planes  are  equal  polygons. 


Given  a  prism  cut  by  two  parallel  planes  forming  the  polygons 
ABODE  and  A'ffC'D'E'. 

To  prove  that  ABODE  =  A'B'C'D'E'. 

Outline  of  Proof :  (1)  Show  that  AB  =  A'B',  BC  =  B'C',  etc., 
by  proving  that  ABB* A',  BCC'B',  etc.,  are  UJ. 

(2)  Show  that  Z  ABC  =  Z  A'B'C',  Z  BCD  =  Z  B'OD1,  etc. 

(3)  Hence  show  that  ABODE  and  A'B'O'D'E'  can  be  made 
to  coincide. 

175.   COROLLARY.     Every  cross-section  of  a  prism  parallel  to 
the  base  is  equal  to  the  base. 

SIGHT  WORK 

1.  Can  a  parallelepiped  be  a  right  prism  without  being  a  rectangular 
parallelepiped  ?    Illustrate. 

2.  Show  that  in  a  rectangular  parallelepiped  each  edge  is  perpendic- 
ular to  the  other  edges  which  meet  it. 

3.  Show  that  any  section  of  a  prism  made  by  a  plane  parallel  to  an  edge 
is  a  parallelogram. 

Suggestion.     Use  in  order  §§  89,  84,  and  92. 

4.  Is  it  possible  to  construct  a  prism  such  that  there  is  no  plane  per- 
pendicular to  its  edges  and  cutting  all  of  them  unless  some  of  them  are 
extended  ? 


56 


-SOLID   GEOMETRY:    BOOK  III 


THE  LATERAL   AREA  OP  A  PRISM 

176.  THEOREM  II.  The  lateral  area  of  a  prism  is 
equal  to  the  product  of  a  lateral  edge  and  the  perimeter 
of  a  right  section. 


Suggestion.  Show  that  the  lateral  edges  are  mutually  equal  and  that 
the  area  of  each  face  is  the  product  of  a  lateral  edge  and  one  side  of 
the  right-section  polygon. 

Complete  the  proof. 

Note.  The  form  of  statement  in  this  theorem  is  the  usual  abbrevia- 
tion for  the  more  precise  form  : 

The  lateral  area  of  a  prism  is  equal  to  the  product  of  the  numerical 
measures  of  a  lateral  edge  and  the  perimeter  of  a  right 
section. 

Similar  abbreviations  are  used  throughout  this  text. 

177,  COROLLARY.  The  lateral  area  of  a  right 
prism  is  equal  to  the  product  of  its  altitude  and  the 
perimeter  of  its  base. 


SIGHT  WORK 

1.  If  the  lateral  edge  of  a  prism  is  8  inches  and  the  perimeter  of  its 
right  section  30  inches,  what  is  the  lateral  area  ? 

2.  Of  all  prisms  having  an  altitude  6  inches  and  the  perimeter  of  a 
right  section  24  inches,  which  one  has  the  smallest  lateral  area  ? 

Suggestion.     Which  one  has  the  shortest  lateral  edge  ? 

3.  Why  is  not  the  lateral  area  of  an  oblique  prism  determined  by  the 
perimeter  of  the  base  and  a  lateral  edge  ? 


PRISMS  AND   CYLINDERS 


57 


CONDITIONS   MAKING  PRISMS   EQUAL 

178.  THEOREM  III.  Two  prisms  are  equal  if  three 
faces  having  a  common  vertex  in  the  one  are  equal 
respectively  to  three  faces  having  a  common  vertex  in 
the  other,  and  similarly  placed. 


ID'. 


Given  the  three  faces  meeting  at  B  in  prism  P  equal  respectively 
to  the  three  faces  meeting  at  B1  in  prism  P1,  and  similarly  placed. 

To  prove  that  P  can  be  made  to  coincide  with  P. 

Outline  of  Proof :    Trihedral  A  B  and  B'  are  equal.        Why  ? 

Now  apply  the  two  prisms,  making  B  coincide  with  B',  and 
then  show  in  detail  that : 

(1)  The  lower  bases  coincide. 

(2)  The  lateral  faces  at  B  and  B'  coincide. 

(3)  The  upper  bases  coincide. 

(4)  All  the  lateral  faces  coincide. 

179,  COROLLARY  1.  Two  truncated  prisms  are  equal  under  the 
conditions  of  §  178. 


180,    COROLLARY  2.     Tivo  right  prisms  are  equal  if  they  have 
equal  bases  and  equal  altitudes. 


58  SOLID   GEOMETRY:    BOOK  III 

OPPOSITE  FACES  OF  A  PARALLELOPIPED 

181,  THEOREM  IV.     Opposite  faces  of  a  parallelo- 
piped  are  equal  and  parallel. 


Suggestions.     Consider  the  opposite  faces  ABFE  and  DCGH. 

(1)  Show  that  the  sides  of  the  angles  ABF  and  DCG  are  parallel,  and 
hence  that  the  planes  determined  by  them  are  parallel. 

(2)  Show  that  these  faces  are  equal. 

In  like  manner  argue  about  any  other  pair  of  opposite  faces. 

182,   COROLLARY.      Any    section    of    a  HR 

parallelopiped    made    by    a   plane   cutting 
four  parallel  edges  is  a  parallelogram. 

Suggestion.     In  the  figure  show  that  EQ  II  ST 
and  QT\\  ES. 

SIGHT  WORK 

1.  A  section  made  by  a  plane  passed  through  two  diagonally  opposite 
edges  of  a  parallelopiped  is  a  parallelogram,  for  example,  the  section 
through  DH  and  BF  in  the  figure  of  §  181. 

2.  Show  that  any  two  of  the  four  diagonals  of  a  parallelopiped  bisect 
each  other. 

Suggestion.     Make  use  of  example  1. 

3.  If  the  parallelopiped  in  §  182  is  rectangular,  is  the  section  neces- 
sarily a  rectangle  ?    May  it  be  so  ? 

4.  Show  that  the  section  of  a  cube  made  by  a  plane  through  two  diago- 
nally opposite  edges  is  a  rectangle.     Why  can  it  not  be  a  square  ?     Can 
you  construct  a  parallelopiped  in  which  such  a  section  would  be  a  square  ? 


PRISMS  AND   CYLINDERS  59 

PLANE  THROUGH  OPPOSITE  EDGES  OF  A  PARALLELOPIPED 

183,  THEOREM  Y.  A  plane  through  diagonally 
opposite  edges  of  a  parallelopiped  divides  it  into  two 
triangular  prisms  having  equal  right  sections  and  equal 
lateral  edges. 


Given  the  parallelopiped  ABCD-F  with  opposite  edges  AE 
and  CG. 

To  prove  that  the  plane  through  AE  and  CG  divides  it  into 
two  triangular  prisms  having  equal  lateral  edges  and  equal  right 
sections. 

Proof :  By  §  182  the  right  section  KLMO  of  the  given  prism 
is  a  parallelogram.  Hence  KM  divides  it  into  two  equal 
triangles.  Hence  the  right  sections  of  the  prisms  ABC-F  and 
ACD-H  are  equal. 

The  lateral  edges  BF  and  DH  are  equal  by  §  168. 

SIGHT  WORK 

1.  If  a  right  section  of  a  prismatic  space  is  a  parallelogram,  what  kind 
of  prism  will  be  cut  out  by  two  right  sections  ? 

2.  If  a  right  section  of  a  prismatic  space  is  a  rectangle,  what  kind  of 
prism  will  be  cut  out  by  two  right  sections  ? 

3.  If  a  right  section  of  a  prismatic  space  is  a  regular  hexagon,  what 
kind  of  prism  will  be  cut  out  by  two  right  sections  ? 

4.  What  kind  of  prism  will  be  cut  out  by  two  right  sections  of  any 
prismatic  space  ? 


60  SOLID  GEOMETRY:    BOOK  III 

EXERCISES 

1.  The  face  angles  of  one  trihedral  angle  of  a  parallelepiped 
are  65°,  70°,  and  75°  respectively.     What  are  the  face  angles  of 
the  other  trihedral  angles  ? 

2.  Show  that  two  prisms   cut  from  the   same   prisma'tic 
space  and  having  equal  lateral  edges  have  equal  lateral  areas. 

3.  Show  that  the  diagonals  of  a  rectangular 
parallelepiped  are  all  equal  to  each  other. 

4.  Show  that  the  square  on  the  diagonal  of 
a  rectangular  parallelepiped  is  equal  to   the 
sum  of  the  squares  on  the  sides  meeting  at  a 
vertex  from  which  it  is  drawn. 

E.g.    in  the  figure  A(?  =  A(?  +  CGZ  =  ABZ  +  BC2  +  CG2. 

5.  Find  the  ratio  of  the  diagonal  to  one  edge  of  a  cube. 

6.  Find  the  edge  of  a  cube  whose  diagonal  is  14  inches. 
Find  the  diagonal  of  a  cube  whose  edge  is  16  inches. 

7.  Find  a  diagonal  of  a  rectangular  parallelepiped  whose 
edges  are  6,  8,  and  10  inches  respectively. 

8.  Are  the  diagonals  of  a  cube  perpendicular  to  each  other  ? 

9.  In  a  rectangular  parallelepiped  the  diagonal  of  the  base 
is  12  in.  and  the  altitude  is  5  in.     Find  the  diagonal  of  the 
parallelopiped. 

10.  If  two  equal  right  prisms  whose  bases  are  equilateral 
triangles  are  placed  together  so  as  to  form  one  prism  whose 
base  is  a  parallelogram,  compare  the  lateral  area  of  the  prism 
so  formed  with  the  sum  of  the  lateral  areas  of  the  original 
prisms. 

11.  A   right   prism   whose   bases   are  regular   hexagons  is 
divided  into  six  prisms  whose  bases  are  equilateral  triangles. 
Compare  the  lateral  area  of  the  original  prism  with  the  sum  of 
the  lateral  areas  of  the  resulting  prisms. 


PBISMS  AND   CYLINDERS 


61 


VOLUME  OP   A   RECTANGULAR  PARALLELOPIPED 

184,  Thus  far  certain  properties  of  prisms  have  been  studied, 
but  no  attempt  has  been  made  to  measure  the  space  occupied  by 
such  a  solid.     For  this  purpose  we  consider  first  a  rectangular 
parallelepiped. 

185,  Numerical  Measure.     In  case  each  edge  of  a  rectangular 
parallelepiped  is  commensurable  with  a  unit  segment,  the  num- 
ber of  times  which  a  unit  cube  is  contained  in  it  is  the  numeri- 
cal measure  or  the  volume  of  the  parallelopiped. 

186.  The  Commensurable  Case.     In  the  com- 
mensurable case  just  described,  the  volume 
is  easily  computed. 

E.g.  if  in  the  figure  one  edge  AC  is  4  units,  and 
an  adjoining  edge  AB  is  3  units,  then  a  cube  as  AK, 
whose  edge  is  one  unit,  may  be  laid  off  4  times  along 
AC  and  a  tier  of  3  •  4  =  12  such  cubes  will  adjoin 
the  face  AD.  Since  the  edge  BE  is  5  units  long,  5 
such  tiers  will  exactly  fill  the  space  within  the  solid.  That  is,  5  •  3  •  4  =  60 
is  the  number  of  cubic  units  in  the  solid. 

Hence  in  this  case 

Volume  =  Length  x  Width  x  Height. 

Again,  if  the  given  dimensions  are  3.4,  2.6,.  4.5  decimeters  respectively, 
then  unit  cubes,  with  edge  one  decimeter,  cannot  be  made  to  fill  exactly 
the  space  inclosed  by  the  figure,  but  cubes  with  edge  each  one  centimeter 
will  do  so,  giving  34,  26,  and  45  respectively  along  the  three  edges  of  the 
figure. 

Therefore,  the  volume  is 

34  .  26  .  45  =  39,780  cubic  centimeters,  or  39.78  cubic  decimeters. 

Hence  in  this  case  also 

Volume  =  Length  x  Width  x  Height. 

187.  Formula  for  Volume  of  a  Rectangular  Solid.     We  therefore 
conclude  that  the  volume  of  a  rectangular  parallelopiped  with 
commensurable  dimensions  is  given  by  the  formula 

Volume  =  Length  x  Width  x  Height. 


62  SOLID   GEOMETRY:    BOOK  III 

VOLUME  OF  A  RECTANGULAR  SOLID 

188,  The  Incommensurable  Case.     If  a  rectangular  parallele- 
piped is  such  that  any  two  of  its  dimensions  are  incommen- 
surable with  each  other,  then  there  is  no  unit  cube,  however 
small,  in  terms  of  which  the  volume  can  be  exactly  expressed. 

But  by  choosing  a  unit  cube  sufficiently  small  we  may  deter- 
•mine  the  volume  of  a  rectangular  solid  which  differs  from  the 
given  one  by  as  little  as  we  please. 

E.g.,  if  the  length  is  5  inches  and  the  width  is  V3  inches,  then  the 
base  cannot  be  exactly  covered  with  equal  cubes,  however  small. 

But  since  V3  =  1.732  '••,  if  we  take  as  a  unit  of  volume  a  cube  one  one- 
hundredth  of  an  inch  on  a  side,  then  the  base  of  a  rectangular  solid  whose 
length  is  5  inches  and  whose  width  is  1.73  inches  can  be  exactly  covered 
with  a  layer  of  these  cubes  and  the  number  of  cubes  in  such  a  layer~ts 
500  x  173  =  86,500. 

If  the  height  is  2  inches,  then  200  layers  will  just  reach  the  top,  making 
500  x  173  x  200  =  17,300,000  cubes. 

And  since  1,000,000  such  small  cubes  make  a  cubic  inch,  the  volume 
of  this  solid  is  17.3  cubic  inches. 

The  portion  of  the  given  solid  not  thus  filled  is  5  inches  long,  2  inches 
high  and  less  than  .003  of  an  inch  in  thickness. 

Hence,  its  volume  is  less  than  5  x  2  x  .003  =  .03  of  a  cubic  inch. 

By  expressing  V3  to  further  places  of  decimals  and  then  using  smaller 
and  smaller  units  of  volume,  successive  rectangular  solids  may  be  found 
which  differ  less  and  less  from  the  given  solid. 

The  foregoing  considerations  constitute  an  informal  proof  of 
the  following  theorem : 

189,  THEOREM  VI.     The    volume 
of  a   rectangular  parallelopiped   is 
equal  to  the  product  of  its   length, 
width,  and  height. 

In  symbols,  V  =  1  •  w  •  h. 

The  above  argument  shows  that  this  theorem  holds  for  all  rectangular 
parallelepipeds  used  in  the  process  of  approximation,  and  hence  it  applies 
to  all  practical  measurements  of  the  volumes  of  such  solids. 


PRISMS  AND   CYLINDERS  63 

190,  COROLLARY  1.    The  volume  of  a  rectangular  parallelo- 
piped  is  equal  to  the  product  of  the  numerical  measures  of  its  base 
and  altitude. 

191,  COROLLARY  2.   If  two  rectangular  parallelopipeds  have 
two  dimensions  respectively  equal  to  each  other,  their  volumes  are 
in  the  same  ratio  as  their  third  dimensions;  and  if  they  have  one 
dimension  the  same  in  each,  their  volumes  are  in  the  same  ratio 
as  the  products  of  their  other  two  dimensions. 

For  if  V  and  V  are  the  volumes,  and  a,  6,  c,  and  a',  6;,  c'  the  dimen- 
sions,  then  Z  =  J^_£  =J?  if  6  =  6- andc  =  c<;  or  ^^  if  „  =  «,». 

192,  Volumes  of  Prisms  in  General.     From  the  formula  for 
volumes  of  rectangular  solids,  V=l'W-h,  we  deduce  the  vol- 
umes of  prisms  in  general  by  means  of  the  principle : 

Two  polyhedrons  have  the  same  volume  if  they  can  be  made  to 
coincide,  or  if  they  can  be  divided  into  parts  which  can  be  made  to 
coincide  in  pairs. 

The  sign  =  between  two  polyhedrons  means  that  they  are  equal  in 
all  respects  ;  that  is,  can  be  made  to  coincide.  The  word  equivalent  is 
used  to  mean  equal  in  volume. 

EXERCISES 

1.  The  dimensions  in  inches  of  a  rectangular  parallelepiped 
are  5,  10,  and  2V2.     Taking  the  approximate  value  of  V2  as 
1.4142,  find  the  approximate  volume.     Show  that  this  differs 
from  the  exact  volume  by  less  than  .001  of  a  cubic  inch,  it  being 
given  that  V2  lies  between  1.41421  and  1.41422. 

2.  If  the  dimensions  in  inches  of  a  rectangular  parallele- 
piped are  5,  4V2,  and  5V3,  find  how  near  an  approximation 
to  the  volume  is  possible  by  taking  V2  between  1.414  and 
1.4143  and  V3  between  1.732  and  1.7321. 

3.  If  the  dimensions  in  inches  of  a  rectangular  parallelepiped 
are  3  V2,  3  V3, 2  V5,  find  the  values  to  be  used  for  V2,  V3,  V5, 
to  obtain  the  volume  within  .001  of  a  cubic  inch. 


64 


SOLID   GEOMETRY:    BOOK  III 


VOLUME  OP  AN  OBLIQUE  PRISM 

193,  THEOREM  VII.  The  volume  of  an  oblique 
prism  is  equal  to  that  of  a  right  prism  having  for  its 
base  a  right  section  of  the  oblique  prism  and  for  its 
altitude  a  lateral  edge  of  the  oblique  prism. 


Given  the  oblique  prism  AD'  with  FGHJK  a  right  section  and 
F'G'fffK'  a  right  section  of  the  prism  extended  so  that  AA'  =  KK'. 

To  prove  that  the  oblique  prism  AD'  has  the  same  volume  as  the 
right  prism  KH1. 

Proof :  In  the  truncated  prisms  AH  and  A'H', 
base  ABODE  =  base  A'B'C'D'E'. 

In  faces  ABFK&nd  A'B'F'K',  we  have 
AB  =  A'B',  Z1=Z3,  Z2  =  Z4, 
and        AK=  A'K'  and  BF=  B'F'. 

.-.  ABFK  and  A'B'F'K'  will  coincide  and  are  equal. 

Likewise,  AKJE  =  A'K'J'E'. 

.-.  AH  =  A'H'.  §  179 

But  the  given  prism       AD'  =  AH-\-  KD', 
and  the  right  prism  KH'  =  A'H'  +  KD'. 

A     T\t  YT'rTt 

Q.  E.  D. 


§  175 

Why? 
Why? 


194,   COROLLARY.     Two   prisms  having  equal  lateral  edges 
and  equal  right  sections  are  equal  in  volume. 


PRISMS  AND   CYLINDERS 


65 


VOLUME  OF   ANY  PARALLELOPIPED 

195,   THEOREM  VIII.     The  volume  of  any  parallel- 
epiped is  equal  to  the  product  of  its  base  and  altitude. 


Given  any  parallelepiped  P  with  area  of  base  b  and  altitude  h. 

To  prove  that  its  volume  is  equal  to  b  x  h. 

Proof :  Considering  face  AK  as  the  base  of  prism  P,  pro- 
duce the  four  edges  parallel  to  AB,  and  lay  off  A'B'  =  AB. 

Through  A'  and  B'  erect  planes  _L  A 'B't  thus  cutting  off 
the  right  prism  Q  with  A'L  as  one  base. 

Now  considering  CL  as  the  base  of  prism  Q,  produce  the 
four  edges  II  CB'  and  lay  off  C'D  ==  CB'. 

At  O  and  D  erect  planes  _L  C'D,  cutting  off  the  right 
prism  jR,  which  is  a  rectangular  parallelepiped  with  base  b". 

Now    prove    in    detail:    (1)    h  =  h'  =  h",    (2)    b  =  b'  =  b", 

(3)  Vol.  P  =  Vol.  Q=Vol.  R,  §  193 

(4)  Vol.  It  =  b"h",  (5)  Vol.  P=  bh. 

196.  COROLLARY.  Two  parallelopipeds  are  equal  in  volume 
if  they  have  equal  altitudes  and  bases  of  equal  areas. 


66  SOLID  GEOMETRY:    BOOK  III 

__  Q _ 

VOLUME  OF  A  TRIANGULAR  PRISM 

197,   THEOREM   IX.     The   volume   of  a    triangular 
prism  is  equal  to  the  product  of  its  base  and  altitude. 


Given  the  triangular  prism  whose  base  is  ABC. 

To  prove  that  the  volume  is  equal  to  the  area  of  A  ABC  mul- 
tiplied by  the  altitude  h. 

Proof :  Complete  the  HJ  ABCD  and  EFQH  and  draw  DH. 
Now  show  that  CDHG  and  ADHE  are  UJ,  and  hence  that 
BH  is  a  parallelepiped.  Use  §  43  (2) 

Let  KLMO  be  a  right  section  of  BH,  and  let  KM  be  the 
line  in  which  the  plane  ACGE  cuts  the  plane  KLMO. 

Then  (1)  prism  ABC-F  =  |  prism  BH,  §§  183,  194 

(2)  But  prism  BH=  h  x  area  of  O  ABCD,  §  196 

(3)  Hence,  prism  ABC-F  =h  x|  area  of  O  ABCD,  that 
is,  prism  ABC-F  =hx  area  of  A  ABC. 

Therefore  the  volume  of  prism  ABC-F  is  equal  to  the  area 
of  its  base  times  its  altitude.  Q.  E.  D. 

SIGHT  WORK 

1.  Find  the  volume  of  a  prism  whose  altitude  is  8  in.  and  whose  base 
is  a  right  triangle  with  legs  5  in.  and  6  in.  respectively. 

2.  Find  the  volume  of  a  right  prism  whose  altitude  is  6  in.  and  whose 
base  is  a  rectangle  with  sides  3  in.  and  5  in. 


PRISMS  AND  CYLINDERS  67 

VOLUME  OF  ANY  PRISM 

198,   THEOREM   X.     The  volume  of  any  prism  is 
equal  to  the  product  of  its  base  and  altitude. 


Outline  of  Proof :  Any  prism  can  be  divided  into  triangular 
prisms  by  planes  passing  through  one  edge  and  each  of  the 
other  non-adjacent  edges. 

The  altitudes  of  the  triangular  prisms  are  the  same  as  that 
of  the  given  prism,  and  the  sum  of  their  bases  is  equal  to  the 
base  of  the  given  prism. 

Now  use  §,  197  and  complete  the  proof. 

199,  COROLLARY  1.     The  product  of  the  base  and  altitude  of 
any  prism  is  equal  to  the  product  of  a  lateral  edge  and  the  area 
of  a  right  section. 

For  each  equals  the  volume  of  the  prism.     See  §§  198,  193. 

200,  COROLLARY  2.     If  two  prisms   have    equivalent   bases, 
their  volumes  are  in  the  same  ratio  as  their  altitudes;  and  if  they 
have  equal  altitudes,  their  volumes  are  in  the  same  ratio  as  the 
areas  of  their  bases. 

SIGHT  WORK 

1.  Find  the  volume  of  a  prism  whose  altitude  is  8  in.  and  the  area  of 
whose  base  is  42  sq.  in. 

2.  The  volume  of  a  prism  is  264  cu.  in.  and  its  altitude  is  8  in.     Find 
the  area  of  its  base. 


68  SOLID  GEOMETRY:   BOOK  III 

EXERCISES 

1.  The  theorem  that  the  volume  of  any  prism  is  equal  to 
the  product  of  its  base  and  altitude  is  of  great  importance. 
What  theorems  of  Book  III  have  been  used  directly  or  indi- 
rectly in  proving  it  ? 

2.  What  dimensions  of  a  prism  must  be  known  in  order  to 
determine  its  lateral  area  by  means  of  the  preceding  theorems 
of  Book  III  ?     What  dimensions  must  be  known  to  determine 
its  volume  ? 

3.  The  edge  of  a  cube  is  e.     Find  the  total  surface  and  the 
volume  in  terms  of  e. 

4.  Find  the  edge  of  a  cube  if  its  total  area  is  equal  numeri- 
cally to  its  volume,  an  inch  being  used  as  the  unit. 

5.  Find  the  volume  of  a  regular  right  triangular  prism 
whose  edges  are  all  equal  to  6  inches. 

6.  Find  the  volume  of  a  regular  right  hexagonal  prism 
whose  edges  are  all  equal  to  10  inches. 

7.  A  side  of  the  base  of  a  regular  right  hexagonal  prism  is 
3  inches.     Find  its  altitude  if  its  volume  is  54  V3  cubic  inches. 
What  is  the  total  area  of  this  prism  ? 

8.  The  volume  of  a  triangular  prism  is  equal  to  the  area  of 
one  lateral  face  multiplied  by  half  the  perpendicular  distance 
of  this  face  from  the  remaining  edge.     Prove. 

9.  The  volume  of  a  regular  right  prism  is  equal  to  the 
lateral   area  multiplied   by   half  the   apothem   of   the   base. 
Prove. 

10.  Prove  that  the  sum  of  the  squares  of  the  four  diagonals 
of  a  rectangular  parallelepiped  is  equal  to  the  sum  of  the 
squares  of  the  twelve  edges  of  the  parallelepiped. 

11.  A  prismatic  space  is  cut  by  two  pairs  of  parallel  planes. 
Prove  that  the  volumes  of  the  two  prisms  so  formed  are  equal 
if  the  two  pairs  of  planes  are  the  same  distance  apart. 


PRISMS  AND   CYLINDERS 


CYLINDERS 

201,  Curved  Surface.     A  surface  no  part  of  which  is  plane  is 
called  a  curved  surface. 

E.g.  the  surface  of  an  eggshell  or  of  a  stovepipe  is  a  curved  surface. 

202,  A  Closed  Plane  Curve.     A  curve  which  can  be  traced  con- 
tinuously by  a  point  moving  in  a 

plane  so  as  to  return  to  it's  original         C\ 
position  without  crossing  its  path      f 
is  a  closed  plane  curve.  -tc^ 

A  closed  plane  curve  is  convex 

if  it  can  be  cut  by  a  straight  line  in  not  more   than   two 
points. 

203,  A  Cylindrical  Surface.    If  a  straight 
line  moves   so  as  to  remain   parallel   to 
itself,  while  it  always  touches  a   closed 
convex  plane  curve  and  is  made  to  trav- 
erse  it   completely,   the   line   is   said   to 
generate  a  closed  convex  cylindrical  surface. 
The  moving  line  is  the  generator,  and  the 
generator  in  any  one  of  its  positions  is  an 
element  of  the  surface. 

204,  Cylinder.     A  solid  bounded  by  a 
cylindrical  surface  and  two  parallel  plane 
sections  cutting  all  its  elements  is  called 
a  cylinder. 

205,  Bases.     Lateral  Surface.     The  two 

parallel  cross-sections  which  bound  a  cyl- 
inder are  its  bases,  and  the  curved  surface 
is  its  lateral  surface. 

206,  Element.     Altitude.     That  part  of   the  generator  of  a 
cylindrical  surface  which  is  included  between  the  bases  of  a 
cylinder  is  called  an  element  of  the  cylinder.     The  altitude  of 
a  cylinder  is  the  perpendicular  distance  between  its  bases. 


70 


SOLID  GEOMETRY :    BOOK  III 


CIRCULAR  CYLINDERS 

207,  Right  Section.     A  right  section  of  a  cylinder  is  made  by 
a  plane  cutting  each  of  its  elements  at  right  angles. 

208.  Circular  Cylinders.     A   circular 
cylinder  is  one  whose  right  section  is 
a  circle. 


The  radius  of  a  circular  cylinder  is 
the  radius  of  its  right  section. 

A  cylinder  whose  elements   are  at 
right  angles  to   its  bases  is  called  a 
right  cylinder.     Otherwise  it  is  an  oblique  cylinder.     A  right 
cylinder  whose  bases  are   circles   is  a  right 
circular  cylinder. 

209,  Axis  of  a  Cylinder.  The  line  passing 
through  the  centers  of  two  right  sections  of 
a  circular  cylinder  is  the  axis  of  the  cylinder. 

A  right  circular  cylinder  may  be  generated  by 
revolving  a  rectangle  about  one  of  its  sides  as  an 
axis.     The  side  opposite  the  axis  generates  the  lateral  surface,  and  the 
sides  adjacent  to  the  axis  generate  the  bases. 


SIGHT  WORK 

1.  What  is  the  locus  of  all  points  in  space  which  are  at  a  perpendicular 
distance  of  6  in.  from  a  straight  line  10  in.  long  ? 

2.  Show  that  any  two  elements  of  a  cylinder  determine  a  plane  sec- 
tion of  the  cylinder  which  cuts  the  two  bases  in  parallel  lines. 

3.  If  every  plane  determined  by  two  elements  of  a.  cylinder  is  perpen- 
dicular to  its  bases,  what  kind  of  cylinder  is  it  ? 

4.  If  one  right  section  of  a  cylinder  is  a  circle,  what  would  appear  to 
be  true  of  all  its  right  sections  ?    For  proof  see  §  213. 

5.  If  a  circular  cylinder  is  oblique,  are  its  bases  circles  ? 

6.  If  an  oblique  cylinder  has  a  circular  base,  is  its  right  section  a 
circle  ? 


PRISMS  AND   CYLINDERS 


71 


LONGITUDINAL  SECTIONS  OP  A  CYLINDER 

210.  THEOREM  XI.  If  a  plane  contains  an  element 
of  a  cylinder  and  meets  it  in  any  other  pointy  then  it 
contains  another  element  also,  and  the  section  is  a 
parallelogram. 


Given  the  cylinder  AC  and  a  plane  M  containing  the  element  AD 
and  another  point  as  P  or  P1. 

To  prove  that  it  contains  another  element  and  that  the  section 
containing  these  two  elements  is  a  parallelogram. 

Proof :  (1)  When  the  point  P  is  in  the  lateral  surface. 

Let  BC  be  the  element  through  P. 

Then  BC  II  AD  and  they  determine  a  plane  M.        §§  203,  73 

Also  AB  II  DC  and  ABCD  is  a  parallelogram.  §  92 

(2)  Wlien  the  point  P  is  in  one  of  the  bases. 

Draw  AB  through  P'.     Then  by  (1)  the  plane  M  contains 
the   element  BC  and    the    section    is    a 
parallelogram. 

211,   Plane  Tangent  to  a  Cylinder.     If  a 

plane  contains  an  element  of  a  cylinder 
and  no  other  point  in  it,  the  plane  is  said 
to  be  tangent  to  the  cylinder,  and  the  ele- 
ment is  called  the  line  of  contact. 

It  may  be  noted  that  a  cross-section  of  a  cylinder  and  its 
tangent  plane  consists  of  a  plane  curve  and  a  line  tangent  to  it. 


72 


SOLID  GEOMETRY:    BOOK  III 


THE  BASES  OP  A  CYLINDER  ARE  EQUAL 

212,   THEOREM  XII.     The  bases  of  a  cylinder  are 
equal  plane  figures. 


Given  a  cylinder  with  the  bases  b  and  b'. 

To  prove  that  b  =  b'. 

Proof :  Take  any  three  points  A,  B,  C  in  the  rim  of  the  base 
b  and  draw  elements  through  these  points,  meeting  the  base  b' 
in  D,  E,  F. 

Show  that  A  ABC  =  A  DEF.  Use  §§  210,  27 

Now,  while  the  elements  AD  and  BE  remain  fixed,  conceive 
CF  to  generate  the  surface  of  the  cylinder. 

Evidently  A  ABC  =  A  DEF  in  every  position  of  CF. 

Hence,  if  base  b'  is  applied  to  base  6  with  these  triangles 
coinciding  in  one  position,  they  will  coincide  in  every  position 
corresponding  to  the  moving  generator. 

That  is,  b'  coincides  with  b.  Q.  B.D. 

The  proposition  of  §  212  may  also  be  stated  in  the  form  of  the  follow- 
ing corollary : 

213,  COROLLARY  1.  Parallel  plane  sections  of  a  cylinder  are 
equal,  if  they  cut  all  the  elements. 


214.   COROLLARY  2.      The  axis  of  a  circular  cylinder  passes 
through  the  center  of  all  its  right  sections. 


PRISMS  AND  CYLINDERS  73 

MEASURING  THE  SURFACE  AND  VOLUME  OF   A  CYLINDER 

215,  Area  of  a  Curved  Surface.      Thus  far  in  geometry  the 
word  area  has  been  used  in  connection  with  plane  figures  only. 
In  some  cases  the  computation  of  an  area  has  been  made  by 
an  approximation  process  only,  as  in  the  case  of  some  rectangles 
and  of  the  circle.     Indeed,  in  these  cases  approximate  meas- 
urement only  is  possible,  since  no  square  unit  exists  in  terms 
of  which  such  areas  can  be  exactly  measured. 

In  the  case  of  any  curved  surface  it  is  evident  that  approxi- 
mate measurement  is  the  only  kind  possible  in  terms  of  a  plane 
area  unit,  since  no  such  unit,  however  small,  can  be  made  to 
coincide  with  a  part  of  such  a  surface. 

216,  Volume  of  a  Solid  Having  a  Curved  Surface.     Since  the 
tmit  of  measure  for  solids  is  the  cube,  it  is  evident  that  a  solid 
having  a  curved  surface  cannot  contain  such  a  unit  an  integral 
number  of  times.     Hence  approximate  measurement  only  of 
the  volumes  of  such  solids  is  possible. 

To  measure  approximately  the  area  and  volume  of  a  cylinder  use  is 
made  of  inscribed  and  circumscribed  prisms. 

217,  Inscribed  Prisms.     A  prism  is  said  to  be  inscribed  in  a 
cylinder  if  its  lateral  edges  are  elements  of  the 

cylinder,   and   their    bases    lie    in    the   same 
planes. 

218,  Circumscribed  Prisms.     A  prism  is  said 
to   be    circumscribed   about    a    cylinder    if   its 
lateral  faces  are  all  tangent  to  the  cylinder, 
and  their  bases  lie  in  the  same  planes. 

It  is  evident  that  by  increasing  the  number  of  lateral  faces  of  the  in- 
scribed or  circumscribed  prisms  the  surface  of  the  latter  may  be  made  to  lie 
as  near  the  surface  of  the  cylinder  as  we  please.  The  lateral  edges  of  the 
prisms  will  remain  equal  to  an  element  of  the  cylinder,  while  the  right 
sections  of  the  prisms  can  be  made  to  differ  as  little  as  we  please  from  the 
right  section  of  the  cylinder. 


74 


SOLID  GEOMETRY:    BOOK  III 


LATERAL  AREA  OF  A  CYLINDER 

219,  Fundamental  Assumption  on  the  Area  and  Volume  of  a 
Cylinder.  We  now  assume  that 

A  cylinder  has  a  definite  lateral  area  and  a  definite  volume 
which  may  be  approximated  as  nearly  as  we  please  by  taking  the 
lateral  areas  and  the  volumes  of  successive  inscribed  or  circum- 
scribed prisms. 


220,  Lateral  Area  of  a  Cylinder.     Since  the  lateral  area  of  a 
prism  is  equal  to  the  product  of  a  lateral  edge  and  the  perimeter 
of  a  right  section  (§  176),  it  follows  that  this  theorem  holds  for 
every  inscribed  or  circumscribed  prism  used  in  approximating 
the  lateral  area  of  a  cylinder. 

The  foregoing  considerations  constitute  an  informal  proof  of 
the  following  theorem : 

221,  THEOREM  XIII.     The  lateral  area 
of  a  cylinder  is  equal  to  the  product  of  an 
element  of  the  cylinder  and  the  perimeter 
of  a  right  section. 

222,  COROLLARY  1.     If  r  is   the  radius  of  a 
right  section  of  a  circular  cylinder,  and  e  is  an  element, 
the  lateral  surface  is  S  =  2  irre  (§  176). 

223,  COROLLARY  2.     If  r  is  the  radius  and  h  is  the 
altitude  of  a  right  circular  cylinder,  then  the  lateral 
surface  is  S  =  2  vrh  (§  177). 


PRISMS  AND   CYLINDERS 


75 


VOLUME  OP   A  CYLINDER 

224,  Since  the  volume  of  a  prism  is  equal  to  the  product  of  its 
altitude  and  the  area  of  its  base,  it  follows  that  this  theorem 
holds  for  every  inscribed  or  circumscribed  prism  used  in  ap- 
proximating the  volume  of  a  cylinder. 

All  practical  computations  of  the  surfaces  and  the  volumes  of  cylinders 
are  approximations  based  upon  the  propositions  stated  in  §§  220,  224. 

The  foregoing  constitutes  an  informal  proof  of  the  following 
theorem  : 

225,  THEOREM  XIV.      The  volume  of  a  cylinder  is 
equal  to  the  product  of  its  altitude  and  the  area  of  its 

base. 

226,  COROLLARY  1.     If  r±  is  the  radius  of  the  right 
section  of  a  circular  cylinder  and  e  is  an  element,  then 
the  volume  is  (§  199)      V  =  ^r\e. 

227,  COROLLARY  2.     If  a  cylinder  of  altitude  h 
has  a  circular  base  whose  radius  is  r2,  then  the  volume 
is 


228,  COROLLARY  3.  If  h  is  the  altitude  and  r  the 
radius  of  the  base  of  a  right  circular  cylinder,  then 
the  volume  is  V  =  irrzh. 

In  this  case  r  =  r\  =  r2,  and  e  =  h,  and  the  formulas  of 
corollaries  1  and  2  become  identical. 

Note.  The  theorems  of  §§  221,  225  were  stated  (and  are  true)  for  any 
cylinders  whatever.  However,  the  theorem  of  §  225  is  available  for  the 
computation  of  the  volume  of  a  cylinder  only  in  case  the  area  of  the  base 
or  of  the  right  section  can  be  computed.  And  this  is  possible  by 
elementary  methods  only  in  case  these  are  circles. 

The  lateral  area  of  a  cylinder  can  be  computed  only  in  case  the  perim- 
eter of  a  right  section  can  be  found,  and  this  is  possible  by  elementary 
methods  only  in  case  the  right  section  is  a  circle. 


76  SOLID   GEOMETEY:    BOOK  III 


SIGHT  WORK 

1.  If  a  prism  is  inscribed  in  a  cylinder,  is  its  lateral  area  greater  or 
less  than  that  of  the  cylinder  ?    Is  its  volume  greater  or  less  than  that  of 
the  cylinder  ? 

2.  If  a  prism  of  say  1000  faces  is  inscribed  in  a  cylinder,  would  the 
theorems  of  §  221  and  §  225  apply  directly  to  this  prism  ?     Would  it  be 
easy  to  detect  experimentally  the  difference  between  this  prism  and  the 
cylinder  ? 

3.  How  would  the  right  section  of  the  prism  and  the  cylinder  of  ex- 
ample 2  be  related  ? 

4.  Discuss  the  questions  in  examples  1,  2,  3  as  related  to  prisms  cir- 
cumscribed about  a  cylinder. 

5.  What  is  the  lateral  area  of  a  cylinder  whose  element  is  8  inches 
and  the  perimeter  of  whose  right  section  is  24  inches  ? 

6.  What  is  the  volume  of  a  cylinder  whose  altitude  is  10  inches  and 
the  area  of  whose  base  is  50  square  inches  ? 

7.  What  is  the  volume  of  a  cylinder  whose  edge  is  4  inches,  and  the 
area  of  whose  right  section  is  24  square  inches  ? 


EXERCISES 

1.  Find  the  volume  of  a  right  circular  cylinder  with  radius 
5  and  altitude  8. 

2.  Find  the  volume  of  a  circular  cylinder  if  the  radius  of 
a  right  section  is  6  and  the  length  of  an  element  is  10. 

3.  Find  the  lateral  area  of  a  cylinder  if  the  perimeter  of  a 
right  section  is  39  and  the  length  of  an  element  is  8. 

4.  Find  the  lateral  area  of  a  right  cylinder  if  the  perimeter 
of  the  base  is  30  and  the  altitude  is  5. 

5.  Find  the  total  surface  area  of  a  right  circular  cylinder  of 
radius  3  and  altitude  5. 

6.  Find  the  volume  of  an  oblique  cylinder  whose  base  is  a 
circle  of  radius  4  and  whose  altitude  is  8. 


PRISMS  AND   CYLINDERS  11 

SUMMARY  OF   BOOK  III 

1.  Define  prismatic  surface  and  prismatic  space. 

2.  Define  prism,  base  of  prism,  lateral  surface,  right  sec- 
tion, right  prism,  parallelepiped,  truncated  prism. 

3.  Give  the  rule  for  finding  the  lateral  area  of  an  oblique 
prism.     How  may  this  rule  be  modified  in  the  case  of  a  right 
prism  ? 

4.  What  can  be  said  about  opposite  faces  of  a  parallele- 
piped as  to  shape  and  size  ? 

5.  Give  two  rules  for  finding  the  volume  of  an  oblique 
prism.     See  §§  193,  198. 

6.  In  proving  the  theorem  on  the  volume  of  a  prism,  we 
considered  triangular  prisms,  general  parallelepipeds,  rectangu- 
lar parallelopipeds,  and  general  prisms.     In  what  order  were 
these  taken  ?     Why  ? 

7.  Show  how  you  would  find  the  approximate  volume  of  a 
rectangular  parallelepiped  whose  dimensions  are  V2,  V3,  V5. 
Find  this  volume  accurately  to  two  places  of  decimals.    Discuss 
the  problem  of  finding  this  volume  correct  to  four  places  of 
decimals.     See  exercises  on  page  63. 

8.  Define  closed  plane  curve,  convex  plane  curve,  cylin- 
drical surface,  cylindrical  space,  cylinder,  right  cylinder,  cir- 
cular cylinder,  tangent  plane  to  a  cylinder. 

9.  Give   the   rule  for  finding  the  lateral  surface  of  any 
cylinder,  of  a  right  circular  cylinder. 

10.  Give  two  rules  for  finding  the  volume  of   an  oblique 
cylinder.     Show  that  these  both  hold  true  for  a  right  circular 
cylinder. 

11.  Describe  what  is  meant  by  the  statement  "  the  area  and 
the  volume  of  a  cylinder  may  be  approximated  by  taking  the 
area  and  the  volume  of  inscribed  or  circumscribed  prisms." 


78  SOLID   GEOMETRY:    BOOK  III 

MISCELLANEOUS  EXERCISES  ON  BOOK  III 

1.  If  the  lateral  surface  of  a  cylinder  and  the  length  of  an 
element  are  known,  can  the  perimeter  of  a  right  section  be 
found  ?     If  the  lateral  area  is  400  TT,  and  an  element  15,  find 
the  perimeter  of  a  right  section. 

2.  The  diameter  of  a  right  circular  cylinder  is  8,  and  the 
diagonal  of  the  largest  rectangle  made  by  a  vertical  section  is 
16.     Find  its  altitude. 

3.  The  volume  of  a  right  circular  cylinder  is  128  TT  cubic 
inches  and  its  altitude  is  equal   to   its   diameter.     Find   the 
altitude  and  the  diameter. 

4.  A  rectangle  whose  sides  are  a  and  b  is  turned  about  the 
side  a  as  an  axis  and  then  about  the  side  b.     Find  the  ratio  of 
the  volumes  of  the  two  cylinders  thus  developed.     Also  find 
the  ratio  of  the  total  surfaces  of  these  cylinders. 

5.  Find  the  diameter  of  a   right   circular   cylinder   if  its 
lateral   area   is  equal  numerically  to  its  volume.     Does   the 
result  depend  upon  the  altitude  of  the  cylinder  ? 

6.  If  the  altitude  of  a  right. circular  cylinder  is  equal  to  its 
diameter,  find  the  ratio  of  the  numerical  values  of  its  total  area 
and  its  volume.     Does  this  depend  on  the  radius  ? 

7.  A  regular  hexagonal  prism  is  inscribed  in  a  right  circular 
cylinder  whose  altitude  is  equal  to  the  diameter.     Find  the 
difference  between  the  volumes  of  the  cylinder  and  the  prism, 
if  a  side  of  the  hexagon  is  4  inches. 

8.  A  cylindrical  tank  8  feet  in  diameter,  partly  filled  with 
water,  is  lying  on  its  side.     If  the  greatest  depth  of  the  water 
is  6  feet,  what  fraction  of  the  volume 

of  the  tank  is  filled  with  water  ? 

9.  In   the   preceding   problem    find 
the  fraction  of  the  volume  occupied  by 
water  if  the  width  of  the  top  of  the 

water  along  a  right  cross-section  of  the  tank  is  4  feet. 


BOOK   IV 


PYRAMIDS  AND  CONES 


229,  Pyramidal  Surface.     Given  a  con- 
vex polygon  and  a  fixed  point  P  not  in 
its  plane.     If  a  line  through,  the  fixed 
point  moves  so  as  to  touch  the  boundary 
of  the  polygon  and  is  made  to  traverse 
it  completely,  the  line  is  said  to  gen- 
erate a  convex  pyramidal  surface. 

The  moving  line  is  the  generator  of  the 
surface,  and  in  any  of  its  positions  it  is 
an  element  of  the  surface.  The  fixed 
point  is  the  vertex  of  the  pyramidal 
surface. 

230,  Nappes.     A   pyramidal    surface   has 
two  parts,  one  on  each  side  of  the  vertex, 
which  are  called  nappes. 

231,  Pyramid.     The  solid  bounded  by  a 
pyramidal  surface  and  a  plane  section  cut- 
ting   all    the    elements,   and    not    passing 
through  the  vertex,  is  called  a  pyramid. 


SIGHT  WORK 

If  a  plane  cuts  both  nappes  of  a  pyramidal  surface,  is  it  possible  that  it 
should  cut  every  element  of  the  surface  ? 

Suggestion.  If  a  plane  M  cuts  both  nappes,  then  a  plane  through  the 
vertex  II  M  meets  the  surface  in  two  elements  which  are  II  M  and  hence 
are  not  cut  by  M . 

-     79 


80  SOLID   GEOMETRY:    BOOK  IV 

CLASSIFICATION  OF  PYRAMIDS 

232,  Faces.     Edges.     The  lateral  surface  of  a  pyr- 
amid   is    composed  of    triangular  faces    having   a 
common  vertex  at  the  vertex  of  the  pyramid,  and 
having  as  bases  the  sides  of  the  polygon  forming 
the  base.     The  sides  common  to  two  such  triangles 
are  the  edges  of  the  pyramid. 

Pyramids  are  classified,  according  to  the  shape  of  the  base,  as 
triangular ',  quadrangular,  pentagonal,  etc. 

233,  Tetrahedron.     A  pyramid  having  a  triangular  base  has 
in  all  four  faces,  and  is  called  a  tetrahedron.     In  this   case 
every  face  is  a  triangle,  and  any  one  of  them  may  be  taken  as 
the  base. 

The  altitude  of  a  pyramid  is  the  perpendicular  distance  from 
the  vertex  to  the  plane  of  the  base. 

234,  Regular  Pyramid.     A  regular  pyramid  is  one  whose  base 
is  a  regular  polygon  such  that  the  perpendicular  from  the  ver- 
tex upon  it  meets  it  at  the  center. 

235,  Properties  of  a  Regular  Pyramid. 

(1)  The  edges  are  equal  to  each  other. 

For  they  cut  off  equal  distances  from  the  foot  of  the  perpendicular. 

(2)  The  faces  are  equal  isosceles  triangles. 

(3)  The  altitudes  of  the  triangular  faces  are  equal  to  each  other. 

236,  Slant  Height.     The  altitude  of  any  one  of  the  triangular 
faces  of  a  regular  pyramid  is  called  its  slant  height. 

SIGHT  WORK 

1.  If  the  slant  height  of  a  regular  pyramid  and  the  apothem  of  its  base 
are  given,   how  may  its  altitude  be  computed  ?    If  the  altitude  and 
apothem  are  given,  how  may  the  slant  height  be  found  ? 

2.  Can  the  expression  "slant  height"  be  applied  to  a  pyramid  if  the 
altitudes  of  its  faces  are  not  all  equal  ?    Discuss  fully. 


PYRAMIDS  AND   CONES 


81 


LATERAL  AREA  OP  A  REGULAR  PYRAMID 

237.  THEOREM  I.  The  lateral  area  of  a  regular 
pyramid  is  equal  to  one  half  the  product  of  its  slant 
height  and  the  perimeter  of  the  base. 


Suggestion  for  Proof  :    Let  I  =  PK  represent  the  slant  height,  S  the 
lateral  area,  and  p  the  perimeter  of  the  base.     We  are  to  show  that 

Of  —   17      *n 

Give  all  the  steps. 

238.  Pyramids  Cut  by  a  Plane.     The  part 
of  a  pyramid  between  its  base  and  a  plane 
cutting  all  its  edges  is  called  a  frustum  if 
the  cutting  plane  is  parallel  to  the  base, 
and  a  truncated  pyramid   if    the   cutting 
plane  is  not  parallel  to  the  base. 

239.  The  Parts  of  a  Frustum.     The  two 
parallel  faces  of  a  frustum  are  its  upper 
and  lower  bases,  the  perpendicular  distance 
between  the  bases  is  its  altitude,  the  trape- 
zoidal faces  are  its  lateral  surface,  and  the 
common  altitude  of  these  faces  is  the  slant 
height  of  the  frustum. 

240.  COROLLARY.     The  lateral  area  of  the  frustum  of  a  right 
pyramid  is  equal  to  half  the  sum  of  the  perimeters  of  the  bases 
multiplied  by  the  slant  height. 

Suggestion.     Use  §  62  and  the  figure  in  §  238, 


82 


SOLID   GEOMETRY:    BOOK  IV 


SECTION  PARALLEL  TO  THE   BASE  OF   A  PYRAMID 

241,    THEOREM  II.     If  a  pyramid  is  cut  ~by  a  plane 
parallel  to  the  base,  then 

(1)  the  edges  and  altitude  are  divided  proportionally  ; 

(2)  the  polygonal  section  is  similar  to  the  base  ; 

(3)  the  areas  of  this  section  and  of  the  base  are  in  the 
same  ratio  as  the  squares  of  their  perpendicular  dis- 
tances from  the  vertex. 


Outline  of  Proof  :  Let  A'B'C'D'E'  be  a  section  II  ABODE. 

'pfT'i       ~PA*       T*TV 
(1)  To  prove  that  =         =         ,  etc.,  pass  another  plane 


through  P  parallel  to  the  base  and  then  use  §  103. 

(2)  To  prove   that  ABODE  ~  A'B'C'D'E',  we  show  that 

Af~Dt  ft'C^ 

Z  A  =  Z.  A1,  Z.B  =  Z  £',  etc.,  and  also          =         ,  etc. 


(3)  Calling  the  area  of  the  section  bf  and  that  of  the  base  b,  we 


are  to  prove  that  -  =          ,  and  for  this  we  need  to  show  that 


A'B'=PA'  = 
AB      PA      PK 

Give  all  the  steps  in  detail. 


PYRAMIDS  AND  CONES 


88 


242.  COROLLARY.  If  two  pyramids  have  equal  altitudes  and 
bases  of  equal  areas  lying  in  the  same  plane,  the  sections  made 
by  a  plane  parallel  to  the  plane  of  the  bases  have  equal  areas. 

Q 


Suggestion.     If   t  and  t'  are  the  areas  of  the  sections  and  b  and  b' 


those  of  the  bases,  show  that-  =  ^-2 ,    -  =  ±-% ,  and  hence  -  =  - ,  from 
b      GT,       b'       PTT  b     b' 


which  it  follows  that  t  =  t'ifb  = 


QL 


SIGHT  WORK 

1.  Find  the  lateral  area  of  a  regular  pyramid  if  its  slant  height  is  10 
in.  and  the  perimeter  of  its  base  is  25  in. 

2.  What  is  the  perimeter  of  the  base  of  a  regular  -pyramid  if  its  lateral 
area  is  72  sq.  in.  and  its  slant  height  is  8  in.  ? 

3.  What  is  the  slant  height  of  a  regular  pyramid  if  its  lateral  area  is 
160  sq.  in.  and  the  perimeter  of  its  base  is  20  in.  ? 

4.  What  is  the  perimeter  of  the  base  of  a  regular  pyramid  if  its  lateral 
area  is  250  sq.  in.  and  its  slant  height  is  20  in.  ? 

5.  The  area  of  the  base  of  a  pyramid  is  36  sq.  in.  and  its  altitude  is 
8  in.     What  is  the  area  of  a  section  of  this  pyramid  made  by  a  plane 
parallel  to  the  base  and  4  in.  from  the  vertex  ? 

6.  The  area  of  the  base  of  a  pyramid  is  64  sq.  in.  and  its  altitude  is 
8  in.    Find  the  distance  from  the  vertex  to  a  plane  parallel  to  the  base  if 
the  area  of  its  section  is  16  sq.  in. 

7.  If  the  area  of  one  section  of  a  pyramid  is  double  that  of  another, 
both  sections  being  parallel  to  the  base,  find  the  ratio  of  their  distances 
from  the  vertex.     Find  this  ratio  if  the  area  of  one  section  is  three  times 
that  of  the  other  ;  also  if  it  is  four  times. 


84 


SOLID  GEOMETRY:    BOOK  IV 


INSCRIBED  AND  CIRCUMSCRIBED  PRISMS 

243,  Construction.     A  triangular  pyramid  is  cut  by  a  series 
of  planes  parallel  to  the  base,  including  one  through  the  ver- 
tex and  also  the  one  in 

which  the  base  lies. 

Through  the  lines  of 
intersection  of  these 
planes  with  one  of  the 
faces,  planes  are  con- 
structed parallel  to  the 
opposite  edge,  thus  form- 
ing a  set  of  prisms  all 
lying  within  the  pyra- 
mid, as  a',  b',  c'  in  pyra- 
mid P',  or  a  set  lying  partly  outside  the  pyramid,  as  a,  6,  c,  d 
in  pyramid  P. 

The  inner  prisms  thus  constructed  are  called  a  set  of  in- 
scribed prisms,  and  the  outer  prisms  are  called  a  set  of  circum- 
scribed prisms. 

This  process  may  be  repeated  by  doubling  the  number  of  planes  drawn 
parallel  to  the  base  and  thus  doubling  the  number  of  inscribed  or  cir- 
cumscribed prisms.  By  continuing  in  this  way  either  set  of  prisms  may 
be  made  to  coincide  as  nearly  as  we  please  with  the  pyramid. 

244.  Fundamental  Assumption  on  the  Volume  of  a  Pyramid. 

We  now  assume  that 

A  pyramid  has  a  definite  volume  which  is  less  than  the  com- 
bined volume  of  any  set  of  circumscribed  prisms  and  greater 
than  that  of  any  set  of  inscribed  prisms. 

SIGHT  WORK 

1.  Show  that  in  the  above  figure  prisms  b  and  a'  are  equal  in  volume 
if  PM=piM'  and  A  ABC  and  A'B'C1  have  equal  areas.     See  §§  188, 
242.     Also  prove  that  volume  c  =  volume  &*,  etc. 

2.  Show  that  the  sum  of  the  volumes  of  the  circumscribed  prisms  ex- 
ceeds the  sum  of  the  volumes  of  the  inscribed  prisms  by  the  volume  of  a. 


PYRAMIDS  AND  CONES  85 

PYRAMIDS  HAVING  EQUAL  VOLUMES 

245.  THEOREM  III.  If  two  triangular  pyramids 
have  equal  altitudes  and  bases  of  equal  areas,  their 
volumes  are  equal. 


B 

Given  the  pyramids  P  and  P1  in  which  altitudes  PM  and  P*M' 
are  equal,  and  the  bases  ABC  and  A'B'C'  have  equal  areas. 

To  prove  that  P  and  P'  have  equal  volumes. 

Proof :  If  P  differs  at  all  in  volume  from  P',  let  P  be  the 
greater,  and  let  the  difference  be  some  fixed  number  K,  so 
that  Vol.  P  -  Vol.  P'  =  K  (1) 

Construct  a  set  of  inscribed  and  circumscribed  prisms  as  in 
§243. 

Then  a'  =  b,  V  =  c,  c'  =  d.  Why  ? 

Denote        a  -f  b  -f-  c  +  d  by  V  and  a'  +  &'  -f  c'  by  F7. 

Then  F-F  =  a.  (2) 

We  have  Vol.  P  <  F,  and  Vol.  P'  >  F'.  §  244 

.-.  Vol.  P  -  Vol.  P'  <  F-  F'.  Why  ? 

Hence  from  (2)        Vol.  P  -  Vol.  P'  <  a. 

Now  take  the  divisions  on  PM  small  enough  to  make  a  <  K. 

§243 

Hence,  Vol.  P  -  Vol.  P  <  K.  (3) 

Thus  (3)  contradicts  (1),  and  hence  the  supposition  that  P 

and  P'  differ  in  volume  is  impossible.  Q.  B.  D. 


86 


SOLID   GEOMETRY:    BOOK  IV 


VOLUME  OF  A   TRIANGULAR  PYRAMID 

246,  THEOREM  IV.  The  volume  of  a  triangular 
pyramid  is  one  third  of  the  product  of  its  base  and 
altitude. 


Given  the  triangular  pyramid  E-ABC.  Let  h,  &,  and  V  be  the 
numerical  measures  respectively  of  the  altitude  EM,  the  area  of  the 
base  ABC,  and  the  volume  of  the  pyramid. 

To  prove  that  V=±bh. 

Proof :  On  the  base  ABC  construct  a  triangular  prism  with 
altitude  h  and  lateral  edge  EB. 

This  prism  may  be  cut  into  three  pyramids  by  the  plane 
sections  through  DEO  and  AEG,  as  shown  in  the  figure  to 
the  right. 

The  pyramids  E-ABC  and  C-DEF  have  equal  volumes 
since  they  have  equal  bases,  ABC  and  DEF  (§  175),  and  the 
same  altitude,  EM. 

Likewise  volume  E-ACD  =  volume  E-CFD. 

But  C-DEF  and  E-CFD  are  the  same  pyramid. 

Hence,  Vol.  E-ABC  =  Vol.  C-DEF  =  Vol.  E-ACD. 

That  is,  Vol.  E-ABC  is  one  third  of  the  volume  of  the 
prism. 

But  volume  of  prism  =  bh.  §  198 

Hence,  V=  volume  of  pyramid  =  £  bh.  Q.  E.  D. 


PYRAMIDS  AND   CONES 


87 


VOLUME  OF   ANY  PYRAMID 

247,  THEOREM  V.     TJie  volume  of  any  pyramid  is 
one  third  of  the  product  of  its  base  and  altitude. 


Given  the  pyramid  P-ABCDE.     Let  V,  b,  and  h  be  the  numeri- 
cal measures  respectively  of  the  volume,  base,  and  altitude. 
To  prove  that  V=^bh. 

Proof :  By  means  of  diagonal  planes  such  as  PAG  and  PAD, 
divide  the  given  pyramid  into  triangular  pyramids.  Call  the 
bases  of  the  triangular  pyramids  bl}  bz,  63,  etc.,  and  their  com- 
mon altitude  h.  Then  by  §  246,  the  volumes  of  the  triangular 
pyramids  are  i.  Ofo  i.  ^  i  5^. 

Hence  the  total  volume  is  ^  (6j  +  62  +  b3)h  =  ^bh.         Q.  E.  D. 

248.  COROLLARY  1.  The  volumes  of  any  two  pyramids  having 
equal  altitudes  are  proportional  to  the  areas  of  their  bases. 


249.   COROLLARY  2.     TJie  volumes  of  any  two  pyramids  having 
bases  of  equal  areas  are  proportional  to  their  altitudes. 


SOLID   GEOMETRY:    BOOK  IV 


VOLUME  OF  A  FRUSTUM  OF  A  PYRAMID 

250.  THEOREM  VI.  The  volume  of  a  frustum  of  a 
pyramid  is  equal  to  the  combined  volumes  of  three 
pyramids  whose  common  altitude  is  the  same  as  that 
of  the  frustum,  and  the  areas  of  whose  bases  are  those 
of  the  upper  and  lower  bases  of  the  frustum  and  the 
mean  proportional  between  these  areas. 

p 


Given  the  frustum  AC  with  area  of  lower  base  &,  area  of  upper 
base  b',  and  altitude  h.  Then  VW  is  the  mean  proportional  be- 
tween b  and  b'. 

To  prove  that  the  volume  of  AC'  is 

V=  %hb  +  %hb'  +  ^hVbb'  =  %h[b  +  b'  -f  V&&7]. 

Proof :  Let  &'  be  the  altitude  PK  of  the  complete  pyramid 
P-ABCDE.  Then  the  altitude  of  the  pyramid  P-A'&C'D'E' 
is  ^'  -  h. 

Hence,  |-=  — - ^——  §  241  (3) 

o       (h  —  h)z 

/r         T/ 
from  which 


Clearing  of  fractions  and  solving  for  h', 

k,  —  ftVft  (1) 


PYRAMIDS  AND   CONES  89 

Now  V  is  the  difference  between  the  pyramids  whose 
altitudes  are  h'  and  h1  —  h. 

Hence,  V=±bh'  -±  b'(h'  -h),  §  247 

or,  rearranging,  V  =  £  b'h  +  £  &'(&  -  6').  (2) 

Substituting  the  value  of  h'  from  (1)  in  (2)  we  have 


Q.E.D. 


SIGHT  WORK 

1.  A  flower  bed  is  in  the  form  of  a  regular  right  pyramid,  with  a 
square  base  5  ft.  on  a  side.     The  altitude  is  3  ft.     Find  the  number  of 
cubic  feet  of  soil  used  in  its  construction. 

2.  The  altitude  of  a  certain  pyramid  is  15  in.  and  its  volume  is  380  cu. 
in.    Find  the  area  of  its  base. 

3.  The   area  of  the  base  of  a  pyramid  is  48  sq.  ft.   and  its  volume 
160  cu.  ft.     Find  its  altitude. 

4.  Find  the  locus  of  the  vertices  of  pyramids  having  the  same  base 
and  equal  volumes. 

5.  Two  monuments  having  bases  of  equal  areas  are  pyramidal  in  shape, 
one  being  15  ft.  high  and  the  other  18  ft.     Find  the  ratio  of  their  volumes. 

6.  Two  pyramids  with  equal  altitudes  have  bases  whose  areas  are 
7  sq.  ft.  and  13  sq.  ft.     Find  the  ratio  of  the  volumes  of  the  pyramids. 

7.  Find  the  volume  of  a  frustum  of  a  pyramid  the  areas  of  whose  bases 
are  36  sq.  in.  and  144  sq.  in.,  and  whose  altitude  is  10  in. 

8.  The  volume  of  a  frustum  of  a  pyramid  is  332  cu.  in.  and  the  areas 
of  its  bases  are  9  sq.  in.  and  36  sq.  in.    Find  its  altitude. 


90  SOLID  GEOMETRY:    BOOK  IV 


EXERCISES  INVOLVING  NUMERICAL  COMPUTATION 

1.  The  base  of  a  regular  pyramid  is  a  square  whose  sides 
are  16  ft.     Find  the  slant  height  of  the  pyramid  if  its  altitude 
is  6  ft.     Also  find  the  lateral  area. 

2.  The   lateral   area   of    a    regular   hexagonal   pyramid   is 
72  sq.  ft.  and  the  slant  height  is  12  ft.     Find  the  perimeter 
of  the  base,  the  apothem  of  the  base,  and  the  altitude  of  the 
pyramid. 

3.  Find  the  lateral  area  of  the  frustum  of  a  pyramid  if  the 
perimeters  of  its  bases  are  27  and  54  in.  respectively  and  its 
slant  height  is  12  in.     Does  the  result  depend  upon  the  num- 
ber of  sides  of  the  frustum  ? 

4.  The  bases  of  a  frustum  of  a  pyramid  are  squares  whose 
sides  are  8  in.  and  2  in.  respectively.     Find  the  volume  of  the 
frustum  if  its  altitude  is  6  in. 

5.  Find  the  volume  of  a  regular  triangular  pyramid  the 
sides  of  whose  base  are  5  in.  and  whose  altitude  is  6  in. 

Suggestion.     The  area  of  a  regular  triangle  with  side  a  is  j-  v/3. 

6.  Find  the  volume  of  a  regular  hexagonal  pyramid  the 
sides  of  whose  base  are  8  in.  and  whose  altitude  is  10  in. 

Suggestion.    The  area  of  a  regular  hexagon  with  side  a  is V3. 

7.  Two  marble  ornaments  of  equal  altitudes  are  pyramidal 
in  form.     One  has  a  square  base  2  in.  on  a  side  and  the  other 
a  regular  hexagonal  base  1  in.  on  a  side.     Find  the  ratio  of 
their  volumes. 

8.  A  pyramid  has  for  its  base  a  right  triangle  with  hypote- 
nuse 10  and  shortest  side  6.     Find  the  volume  of  the  pyramid 
if  its  altitude  is  9. 

9.  The  slant  height  of  a  frustum  of  a  regular  pyramid  is 
10  in.  and  the  apothems  of  its  bases  are  8  in.  and  6  in.  re- 
spectively.    Find  its  altitude. 


PYRAMIDS  AND   CONES  91 

EXERCISES  INVOLVING  ALGEBRAIC  COMPUTATION 

1.  A  tent  is  to  be  made  in  the  form  of  a  right  pyramid 
with  a  regular  hexagonal  base.     If   the   altitude  is  fixed  at 
12  ft.,  what  must  be  the  side  of  the  base  in  order  that  the 
tent  may  inclose  400  cu.  ft.  of  space  ? 

2.  A  pyramid  with  altitude  8  in.  and  a  base  whose  area  is 
36  sq.  in.  is  cut  by  a  plane  parallel  to  the  base  so  that  the 
area  of  the  section  is  18  sq.  in.     Find  the  distance  from  the 
base  to  the  cutting  plane. 

3.  A  frustum  of  a  pyramid  is   cut   from   a   pyramid   the 
perimeter  of  whose  base  is  60  in.  and  whose  altitude  is  15  in. 
What  is  the  altitude  of  the  frustum  if  the  perimeter  of  its 
upper  base  is  20  in.  ?     Does  the  result  depend  upon  the  num- 
ber of  sides  of  the  pyramid  ? 

4.  Solve  the  preceding  problem  if   the   perimeter   of  the 
upper  base  of  the  frustum  is  one  nth  that  of  the  lower  base. 

5.  The  area  of  the  base  of  a  pyramid  is  180  sq.  in.  and  its 
altitude  is  20  in.     Cut  from  it  a  frustum  the  area  of  whose 
upper  base  is  45  sq.  in. ;  also  one  the  area  of  whose  upper  base 
is  one  nth  of  180  sq.  in.     Do  these  results  depend  upon  the 
number  of  sides  of  the  pyramid? 

6.  If  the  altitude  of  a  pyramid  is  h,  how  far  from  the  base 
must  a  plane  parallel  to  it  be  drawn  so  that  the  area  of  its 
cross-section  shall  be  half  that  of  the  base  of  the  pyramid  ? 

7.  In  a  regular  right  pyramid  a  plane  parallel  to  the  base 
cuts  it  so  as  to  make  a  section  whose  area  is  one  half  that  of 
the  base.     Find  the  ratio  between  the  lateral  area  of  the  pyra- 
mid and  that  of  the  small  pyramid  cut  off  by  the  plane. 

8.  Find  the  volume  and  the  total  surface  of  a  regular  tetra- 
hedron whose  edges  are  9  in. 

9.  Find  the  total  surface  and  the  volume  of  a  regular  hexag- 
onal pyramid  the  sides  of  whose  base  are  each  a  and  whose 
altitude  is  a. 


92 


SOLID  GEOMETRY:    BOOK  IV 


CONES 

251,  Conical  Surface.     Given  a  closed  convex 
plane  curve  and  a  fixed  point  P  not  in  its 
plane.     If    a   line   through   P  moves    so    as 
always  to  touch  the  curve   and   is   made  to 
traverse  it  completely,  it  is  said  to  generate 
a  convex  conical  surface. 

The  moving  line  is  the  generator  of  the  sur- 
face, and  in  any  of  its  positions  it  is  an  element 
of  the  surface.  The  fixed  point  is  the  vertex. 

252,  Nappes.     A    conical    surface    has    two 

parts,  one  on  each  side  of  the  vertex,  which  are  called  nappes. 

253,  Cone.     The  solid  bounded  by  a  conical  surface  and  a 
plane  section  cutting  all  its  elements,  and 

not  passing  through  the  vertex,  is  called  a 
cone. 

254,  Base.  Lateral  Surface.  Altitude.    The 
plane  part  of  the  surface  of  a  cone  is  its 
base  and  the  curved  part  is  the  lateral  sur- 
face.    The  altitude  of  a  cone  is  the  perpen- 
dicular distance  from  the  vertex  to  the  plane 
of  the  base. 

255,  Circular  Cone.    A  cone  which  has  a  circular  cross-section 
such  that  the  perpendicular  upon  it  from  the  vertex  meets  it 
at  the  center  is  called  a  circular  cone.     If  the  base  is  such  a 
circle,  the  cone  is  then  a  right  circular  cone.     Otherwise,  it  is 
an  oblique  circular  cone. 

The   axis   of  a  circular  cone  is  the  line  from  the  vertex 
through  the  center  of  a  circular  section. 

SIGHT  WORK 

If  a  plane  cuts  both  nappes  of  a  conical  surface,  show  that  it  cannot 
cut  all  the  elements  of  the  surface.     See  suggestion  on  page  79. 


PYRAMIDS  AND  CONES 


93 


256,  Generating  a  Right  Circular  Cone.     A  right  circular  cone 
may  be  generated  by  rotating  a  right  triangle  PMB  about  one 
of  its  legs,  PJf,  as  an  axis.     The  hypotenuse  PB  generates  the 
lateral  surface,  and  the  other  leg,  MB, 

generates  the  base. 

257,  Slant  Height.     The   generator 
of  the  convex  surface  of  a  right  cir- 
cular cone  in  any  of  its  positions  is 
called  the  slant  height  of  the  cone. 

258,  Frustum  of  a  Cone.     The  part  of  a  cone  included  be- 
tween   the    base    and    a   plane 

section  parallel  to  the  base  is 
called  a  frustum  of  a  cone. 

The  base  of  the  cone  and  the 
parallel  section  are  the  bases  of 
the  frustum. 

259,  Slant  Height  of  a  Frustum.     A  frustum  cut  from  a  right 
circular  cone  has  two  circular  bases.     The  segments  of  all  the 
elements  of  the  cone  intercepted  between  these  bases  are  equal 
and  their  common  length  is  the  slant  height  of  the  frustum. 


SIGHT  WOIJK 

1.  Compare  the  definitions  of  pyramidal  and  conical  surfaces. 

2.  Compare  the  definitions  of  a  pyramid  and  a  cone,  and  of  a  frustum 
of  each. 

3.  What  kind  of  cone  corresponds  to  a  regular  pyramid  ? 

4.  Can  the  expression  "slant  height "  be  applied  to  any  other  cone 


than  a  right  circular  cone  ? 
cone. 


Discuss  this  question  for  the  frustum  of  a 


5.  In  a  right  circular  cone,  if  any  two  of  the  three  quantities,  altitude, 
slant  height,  radius  of  the  base,  are  given,  show  that  the  third  may  be 
found. 


94 


SOLID   GEOMETRY:    BOOK  IV 


SECTION  OF  A  CONE  THROUGH  AN  ELEMENT 

260,  THEOREM  VII.  If  a  plane  contains  an  element 
of  a  cone  and  meets  its  surface  in  any  other  point,  then 
it  contains  another  element  also,  and  the  section  is  a 
triangle. 


Let  a  plane  contain  the  element  PB  of  the  cone  P-ABC,  and  also 
one  other  point  K  or  K'. 

To  prove  that  this  plane  contains  another  element  PD,  and  that 
the  section  is  a  triangle  PBD. 

Outline  of  Proof :  (1)    When  the  point  K  is  in  the  lateral  surface. 

Draw  the  element  PD  through.  K,  and  let  the  plane  deter- 
mined by  PB  and  PD  meet  the  base  in  BD. 

Then  A  PBD  is  the  section  made  by  the  plane  containing  PB 
and  the  point  K. 

(2)    When  the  point  1C  is  in  the  base. 

Draw  BD  through  K'  and  also  the  element  PD. 

Then  the  plane  determined  by  BD 
and  PD  contains  PB  and  K'  and  cuts  the 
cone  in  the  A  PBD. 

261,  Plane  Tangent  to  a  Cone.  If  a 
plane  contains  an  element  of  a  cone  and  no 
other  point  of  the  cone,  the  plane  is  tan- 
gent to  the  cone,  and  the  element  is  called 
the  element  of  contact. 


PYRAMIDS  AND   CONES 


95 


SECTION  OF   A  CONE  PARALLEL   TO  THE  BASE 

262,  THEOKEM  VIII.     If  the  base  of  a  cone  is  circular, 
every  plane  section  parallel  to  the  base  is  also  circular. 


Given  a  cone  with  a  circular  base  AD. 

To  prove  that  the  \\  section  EH  is  also  circular. 

Proof :  Draw  the  straight  line  from  P  to  the  center  M  of  the 
base,  and  let  it  meet  the  section  EH  in  the  point  0.  Let  F 
and  G  be  any  two  points  on  the  perimeter  of  the  section  EH. 

Pass  planes  containing  PM  through  the  points  F  and  G,  and 
let  them  cut  the  base  in  MB  and  MC  respectively. 

Now  in  A  PMB  and  PMC,  OF  II  MB  and  OG  II  MC.      §  92 


But 


OF^OP  =  OG 
'  MB     MP     MC 

MB  =  MC.     .'.  OF=  OG. 


Why? 


Hence,  since  F  and  G,  any  two  points  on  the  perimeter  of 
this  section,  are  equally  distant  from  O,  this  shows  that  EH 
is  a  circle  whose  center  is  0. 


Q.  E.  D. 


263,  COROLLARY.  If  a  cone  has  a  circular  base,  the  areas  of 
two  cross-sections  parallel  to  it  are  in  the  same  ratio  as  the  squares 
of  their  perpendicular  distances  from  the  vertex  and  also  as  the 
squares  of  the  distances  of  their  centers  from  the  vertex. 


96     '  SOLID   GEOMETRY:    BOOK  IV 


INSCRIBED   AND   CIRCUMSCRIBED  PYRAMIDS 

264,  The  lateral  surface  of  a  regular  pyramid  inscribed  in  a 
right  circular  cone  may  be  computed  and  is  equal  to  one  half 
the  product  of  the  slant  height  and  the  perimeter 

of  the  base  (§  237). 

If  the  number  of  faces  of  the  inscribed  pyramid 
is  doubled,  the  lateral  surface  of  the  resulting 
pyramid  may  again  be  computed  in  terms  of  its 
slant  height  and  the  perimeter  of  its  base. 

In  a  similar  manner,  the  lateral  surface  of  a 
regular  pyramid  circumscribed  about  a  right 
circular  cone  may  be  computed  in  terms  of  the 
slant  height  of  the  cone  and  the  perimeter  of 
the  polygon  circumscribed  about  the  base. 
The  number  of  faces  may  be  doubled  and  the 
lateral  surface  again  computed,  and  so  on. 

Evidently  either  of  these  processes  may  be  repeated  indef- 
initely and  the  surfaces  of  the  inscribed  or  circumscribed 
pyramids  may  be  made  to  lie  as  close  to  the  surface  of  the  cone 
as  we  please. 

The  circumscribed  pyramids  all  have  the  same  slant  height  as  that  of 
the  cone,  and  in  case  of  the  inscribed  pyramids,  the  slant  height  may  be 
made  to  differ  by  as  little  as  we  please  from  that  of  the  cone  by  making 
the  number  of  faces  great  enough. 

265,  Fundamental  Assumption  on  the  Lateral  Area  of  a  Right 
Circular  Cone.     We  now  assume  that 

A  right  circular  cone  has  a  definite  lateral  area  which  can  be 
approximated  as  nearly  as  we  please  by  taking  the  lateral  area  of 
the  successive  inscribed  or  circumscribed  pyramids. 

Since,  by  the  theorem  of  §  237,  the  lateral  area  of  any  regular  pyramid 
is  half  the  product  of  the  perimeter  of  its  base  and  its  slant  height,  it 
follows  that  this  theorem  holds  for  all  pyramids  used  in  approximating 
the  lateral  area  of  a  right  circular  cone  and  that  all  practical  measure- 
ments of  such  lateral  areas  are  based  on  this  theorem. 


PYRAMIDS  AND  CONES  97 

LATERAL  AREA  OF   A  RIGHT   CIRCULAR  CONE 

The  foregoing  considerations  constitute  an  informal  proof 
of  the  following  theorem  : 

266.  THEOREM  IX.     The  area  of  the  lateral  surface 
of  a  right  circular  cone  is  equal  to  one  half  the  product 
of  its  slant  height  and  the  circumference  of  its  base. 

The  argument  used  on  page  96  shows  that  a  theorem  of  this 
kind  holds  for  every  inscribed  or  circumscribed  pyramid  used  in 
the  approximation  process,  and  hence  this  theorem  for  the  cone 
is  established  for  all  purposes  of  practical  measurement. 

267.  COROLLARY.     If  I  is  the  slant  height 
of  a  right  circular  cone  and  r  is  the  radius  of 
its  base,  the  area  of  the  lateral  surface  is 


Note.  In  the  case  of  a  cone  which  is  not  a  right  circular  cone  the 
slant  height  varies  from  point  to  point  and  the  process  of  computation  of 
§  266  fails.  Finding  the  lateral  surface  of  such  a  cone  depends  on 
methods  first  introduced  in  the  calculus  and  is  a  much  more  difficult 
problem  than  those  solved  in  elementary  plane  and  solid  geometry. 

SIGHT  WORK 

1.  If  a  pyramid  is  inscribed  in  a  cone,  is  its  lateral  area  greater  or  less 
than  that  of  the  cone  ? 

2.  Does  the  theorem  of  §  237  apply  to  an  irregular  pyramid  or  to  an 
oblique  pyramid  ?     Discuss  fully. 

3.  If  a  regular  pyramid  of  1000  faces  is  inscribed  in  a  right  circular 
cone,  would  the  theorem  of  §  237  apply  to  this  pyramid  ?     Would  it  be 
easy  to  detect  experimentally  the  difference  between  this  pyramid  and  the 
cone  ?    Would  the  size  of  the  cone  affect  the  answer  to  this  question  ? 
Discuss  similar  questions  about  a  circumscribed  pyramid  of  1000  faces. 

4.  Find  the  lateral  surface  of  a  right  circular  cone  whose  altitude  is 
7  in.  and  the  radius  of  whose  base  is  2  in. 


98 


8OLTD   GEOMETRY:    BOOK  IV 


LATERAL  AREA  OF  A  FRUSTUM 

268,  THEOREM  X.  The  lateral  area  of  a  frustum 
of  a  right  circular  cone  is  equal  to  one  half  the  sum 
of  the  circumferences  of  the  bases  multiplied  ~by  the, 
slant  height. 


Given  the  frustum  ABCD,  with  slant  height  /  and  radii  of  bases 
r  and  r'. 

Let  S  represent  its  lateral  area. 
To  prove  that  S  =  |(2  TTT  +  2  ,rr')J  =  Trl(r  +  r'). 
Proof :  Complete  the  cone,  and  let  PC  =  V- 
Then  S  =  |  [2  TTT  (I  + 1*)  - 


But 


1  +  l'}  from  which  V  =    r>l 


r'        I 
Substituting  V  from  (2)  in  (1), 

S  =  irrl  +  vr'l  =  7rl(r-4-  r'). 


r  —  r' 


§267 
(1) 
(2) 

Q.  E.  D. 


269,  COROLLARY.  TJie  lateral  area  of  a  frustum  of  a  right 
circular  cone  is  equal  to  the  circumference  of  a  section  midway 
between  the  bases  multiplied  by  the  slant  height. 

Suggestion.     In  the  formula  of  the  theorem  we  have 


S  =  TT  I  (  r  +  r  '  )  =  2  TT  iL±l!    l  . 


Now  show  that 
two 


r  + 


is  the  radius  of  the  section  midway  between  the 

§46 


PYRAMIDS  AND   CONES  99 

SIGHT  WORK 

1.  If  a  triangle  which  is  not  a  right  triangle  revolves  about  one  of  its 
sides,  does  it  generate  a  cone  ? 

Show  that  the  figure  generated  by  revolving  any  triangle  about  its 
longest  side  may  be  divided  into  two  cones. 

2.  Find  the  lateral  area  of  a  right  circular  cone  with  radius  of  base  6  in. 
and  slant  height  10  in.     In  this  and  the  succeeding  exercises,  express  the 
results  in  terms  of  TT,  e.gr.,  60  ?r  sq.  in. 

3.  Find  the  lateral  area  of  a  right  circular  cone  with  radius  of  base  8  in. 
and  altitude  6  in. 

4.  Find  the  lateral  area  of  a  frustum  of  a  right  circular  cone,  the 
radii  of  whose  bases  are  8  in.  and  4  in.  and  whose  slant  height  is  6  in. 

5.  A  right  circular  cone  having  a  base  with  radius  6  ft.  and  altitude 
8  ft.  is  cut  by  a  plane  parallel  to  its  base  and  at  a  distance  of  4  ft.  from 
the  vertex.     Find  the  radius  of  this  section.     Also  find  its  area. 


EXERCISES 

1.  The  lateral  area  of  the  surface  of  a  right  circular  cone  is 
120  TT  sq.  in.,  and  its  radius  is  4  in.     Find  its  slant  height. 

2.  A  circular  chimney  100  ft.  high   is   in   the   form   of  a 
frustum  of  a  right  circular  cone  whose  lower  base  is  10  ft.  in 
diameter  and  upper  base  8  ft.     Find  the  lateral  surface. 

3.  The  lateral  area  of  a  frustum  of  a  right  circular  cone  is 
60  TT  sq.  in. ;  the  radii  of  the  two  bases  are  6  in.  and  4  in.    Find 
the  slant  height  of  the  frustum. 

4.  The  lateral  area  of  a  right  circular  cone  is  S,  and  the  slant 
height  is  I.     Express  the  radius  of  the  base  and  also  the  alti- 
tude in  terms  of  S  and  L 

5.  If  the  radius  of  the  base  of  a  right  circular  cone  is  r,  and 
the  lateral  area  is  S,  express  the  slant  height  in  terms  of  r 
andtf. 

6.  If  the  slant  height  of  a  right  circular  cone  is  Z,  and  the 
lateral   area  is  S,  express  the  circumference  of  the  base   in 
terms  of  I  and  S. 


100 


SOLID   GEOMETRY:    BOOK  IV 


VOLUME  OF  A  CONE 

270,  Consider  inscribed  and  circumscribed  pyramids  similar 
to  those  used  in  §  264,  except  that  they  need  not  be  regular 
since  the  cone  is  not  now  required  to  be  a  right  circular  cone. 

By  repeatedly  increasing  the  number  of  faces  of  either  the 
inscribed  or  circumscribed  pyramids,  they  may  be  made  to 
approach  coincidence  with  the  cone  as  nearly  as  we  please. 

271,  Fundamental  Assumption  on  the  Volume  of  a  Cone.     We 
now  assume  that  a  cone  has.  a  definite  volume  which  can  be  ap- 
proximated as  nearly  as  we  please  by  taking  the  volume  of  succes- 
sive inscribed  or  circumscribed  prisms. 

Recalling  that  the  volume  of  a  pyramid  is  equal  to  one 
third  of  the  product  of  its  base  and  altitude,  we  see  that  the 
above  considerations  constitute  an  informal  proof  of  the 
following  theorem : 


272.  THEOREM  XI.     The  volume   of  a 
cone  is  equal  to  one  third  of  the  product 
of  its  base  and  altitude. 

273,  COROLLARY.     If  the  base  of  a  cone  is  a 
circle  with  radius  r  and  if  the  altitude  is  h,  then 
the  volume  of  the  cone  is 


SIGHT  WORK 

1.  Find  the  volume  of  an  oblique  cone  with  altitude  8  in.  and  a  cir- 
cular base  whose  radius  is  6  in. 

2.  Find  the  volume  of  a  right  circular  cone  with  slant  height  10  in. 
and  radius  of  base  6  in. 

3.  The  area  of  the  base  of  a  cone  is  50  sq.  in.  and  its  volume  is  600 
cu.  in.     Is  the  altitude  the  same  whether  the  cone  is  right  or  oblique  ? 

4.  Show  that  if  two  cones  have  bases  of  equal  areas  their  volumes  are 
proportional  to  their  altitudes. 


PYRAMIDS  AND   CONES 


101 


VOLUME  OF  A  FRUSTUM  OF  A  CONE 

274,  THEOREM  XII.  The  volume  of  the  frustum  of 
a  cone  is  equal  to  the  combined  volumes  of  three  cones 
whose  common  altitude  is  the  altitude  of  the  frustum 
and  whose  bases  are  the  upper  and  lower  bases  of  the 
frustum  and  a  mean  proportional  between  these  bases. 


Suggestion.  The  proof  is  exactly  like  that  of  §  250,  making  use  of 
§  272,  instead  of  §  247.  The  result  in  symbols  is  V=  $  h(b  +  b>  +  Vbb'). 

275,  COROLLARY.  The  volume  of  a  frustum  of  a  right  cir- 
cular cone  is  V—  \  irh(r2  +  r'2  +  rr'),  where  h  is  the  altitude  and 
r  and  r'  are  the  radii  of  the  bases. 

For  b  =  irr*,  b'  =  irr'*.     .'.  V=%  U(TTI*  +  irr'2  +  Virr*  -  7rr'2) 


Note.  The  theorem  of  §  272  holds  for  any  cone  whatever,  whether 
right  or  oblique.  However,  when  the  base  is  not  a  circle  we  have  no 
means  in  elementary  mathematics  for  computing  its  area.  Hence  the 
volume  of  such  a  cone  cannot  be  found  at  this  stage  even  though  §  272 
does  apply. 

Similarly  §  274  applies  to  a  frustum  of  any  cone  whatever,  but  we  are 
able  to  compute  its  volume  by  elementary  methods  only  in  case  the  bases 
are  circles. 

SIGHT  WORK 

The  radius  of  the  base  of  a  cone  is  5  in.  and  its  altitude  is  10  in.  Find 
the  volume  of  a  frustum  formed  by  a  plane  parallel  to  the  base  and  6  in. 
from  it.  Find  the  total  surface  of  this  frustum  in  case  the  cone  is  a  right 
circular  cone. 


102  SOLID   GEOMETRY:    BOOK  IV 

SUMMARY  OP  BOOK  IV 

1.  Define  pyramidal  surface  and  conical  surface.     In  what 
respects  do  they  differ  ? 

2.  Define   pyramid,  cone,  regular  pyramid,  circular  cone, 
and  right  circular  cone. 

3.  For  what  kind  of  cones  may  the  lateral  area  be  found 
by  means  given  in  this  Book?     What  is  the   corresponding 
kind  of  pyramid? 

4.  For  what  kind  of  frustum  of  a  cone  may   the   lateral 
area  be  found  by  means  given  in  this  Book?     What  is  the 
corresponding  kind  of  frustum  of  a  pyramid  ? 

5.  What   assumption   about  the   volume  of  a  pyramid  is 
made  in  Book  IV?     In  the  proof  of  what  theorems  is  this 
assumption  used  ? 

6.  Beginning  with  the  theorem  of  §  245  state  in  order  the 
theorems  which  lead  to  the  rule  for  finding  the  volume  of  any 
pyramid. 

7.  What  assumption  about  the  lateral  area  of  a  cone   is 
made  in  this  Book  ?     Compare  this  assumption  with  the  one 
in  §  219. 

8.  What  assumption  is  made  about  the  volume  of  a  cone  ? 
Compare  this  with  the  assumption  in  §  219. 

9.  What   theorems    on   cylinders   have   no    corresponding 
theorems  for  cones  ? 

10.  Show  that  a  frustum  of  a  cone  becomes  more  and  more 
nearly  identical  with  a  cylinder  if  the  vertex  of  the  cone  is 
removed  farther  and  farther  from  the  base. 

11.  If  a  cone  were  regarded  as  a  pyramid  with  a  very  large 
number  of  very  narrow  lateral  faces,  what  pages  in  this  Book 
would  be  superfluous  ? 

12.  Describe  the  difficulty  in  finding  the  lateral  area  of  an 
oblique  cone  or  of  a  cone  whose  base  is  not  circular.     Does  the 
same  difficulty  exist  in  finding  the  volume  of  such  a  cone  ? 


PYRAMIDS  AND  CONES 


103 


EXERCISES   ON  BOOK  IV 

1.  If  several  planes  are  tangent  to  the  same  cone,  find  one 
point  common  to  them  all. 

2.  Find  the  locus  of  all  lines  which  make  a  given  angle  with 
a  given  line  at  a  given  point  in  it. 

3.  Find  the  locus  of  all  lines  which  make 
a  given  angle  with  a  given  plane  at  a  given 
point. 

4.  If  the  middle  points  of  four  edges  of 
a  tetrahedron,  no  three  of  which  meet  at  the 
same  vertex,  are  joined,  prove  that  a  parallelo- 
gram is  formed. 

5.  Show  how  to  pass  a  plane  through  a 
tetrahedron  so  that  the   section   shall   be   a 
parallelogram. 

Suggestion.    Pass  a  plane  parallel  to  each  of  two 
opposite  edges.     See  §  133. 

6.  A  mound  of  earth  of  the  shape  shown  in  the  figure  has 
a  rectangular  base  16  yards  long  and  8 

yards  wide.     Its   perpendicular   height 
is   5  yards,  and   the  length  on  top   is 


cubic 


8   yards.     Find   the   number    of 
yards  of  earth  in  the  mound. 

Suggestion.    If  from  each  end  a  pyramid  with  a  base  8  yd.  by  4  yd. 
is  removed,  the  remaining  part  is  a  triangular  prism. 

7.  Given  a  figure  in   general  shape   the  same  as  the  pre- 
ceding, with  a  rectangular  base  of  length  24  ft.  and  width  6 
ft.     Find  its  volume  and  lateral  area  if  the  dihedral  angles 
around  the  base  are  each  45°. 

8.  Find  the  area  and  volume  of  the  figure  developed  by  an 
equilateral  triangle  with  sides  a  if  it  is  revolved  about  one  of 
its  sides. 

Suggestion.     The  figure  may  be  divided  into  two  cones. 


104 


SOLID   GEOMETRY :    BOOK  IV 


9.   Find  the  volume  and  area  of  the  figure  formed  by  re- 
volving an  equilateral  triangle  with  sides  s  about  an  altitude. 

10.  Find  the  area  and  volume  of  the  figure  de- 
veloped  by  revolving   a   square  whose   side  is  a 
about  one  of  its  diagonals. 

11.  Through  one  vertex  of  an  equilateral  tri- 
angle with  sides  a  draw  a  line  I  perpendicular  to 
the   altitude   upon  the  opposite   side.     Find   the 
volume  and  area  of  the  figure  developed  by  revolv- 
ing the  triangle  about  the  line  I. 

Suggestion.     The  volume  may  be  obtained  by  subtracting  the  volumes 
of  two  cones  from  the  volume  of  a  cylinder. 


12.  Through  a  vertex  of  a  square  with  sides  a 
draw  a  line  I  perpendicular  to  the  diagonal  at  that 
point.     Find  the  area  and  volume   of   the  figure 
developed  by  turning  the  square  around  Z. 

Suggestion.     Notice  that  two  frustums  and  two  cones  are  developed. 

13.  In  a  regular  hexagon  with  sides  a  draw  a  line 
I  bisecting  two  opposite  sides.     Find  the  area  and 
volume  of  the  figure  developed  by  turning  the  hex- 
agon about  I  as  an  axis. 

14.  One  angle  of  a  right  triangle  is  30°.     Find 
the  ratios  of  the  surfaces  and  also  of  the  volumes 

of  the  solids  developed  by  revolving  this  triangle  around  each 
of  its  three  sides  in  succession.  & 

Suggestion.    The  sides  of  the  A  are  a,  2  a 


15.   If  through  any  point  P  in  a  diag- 
onal of  a  parallelepiped  planes  KN  and 
RM  are   drawn   parallel   to   two  faces, 
show  that  the  parallelepipeds  DQ  and  LN  thus  formed 
equal  volumes. 


B 


have 


Archimedes  (287-212  B.C.)  was  without  doubt  the  greatest  mathema- 
tician of  his  time.  He  is  known  not  through  any  extended  treatise, 
like  that  of  Euclid,  but  through  a  series  of  monographs  on  practi- 
cally every  mathematical  subject  then  known,  including  physics,  me- 
chanics, astronomy,  and  many  phases  of  geometry.  His  work  on 
the  circle,  cone,  cylinder,  and  sphere  is  reflected  in  our  treatment  of 
surfaces  and  volumes  at  the  present  day. 


BOOK   V 


THE  SPHERE 

276,  Sphere,    Center.     A  sphere  is  a  solid  bounded  by  a  sur- 
face all   points  of   which  are   equally  distant   from   a   point 
within  called  the  center. 

277,  Diameter.      Radius.      A   line-segment    connecting   two 
points   on   the   surface  of  a  sphere  and   passing  through  its 
center  is  a  diameter.     A  segment  joining  the  center  to  a  point 
on  the  surface  is  a  radius. 

278,  Notation  for  a  Sphere.     A  sphere 
may  be  designated  by  a  single  letter  at 
the   center  or  by  two  letters  giving  a 
radius. 

Thus  the  sphere  C  means  one  whose  center 
is  C,  and  the  sphere  CA  is  one  whose  center 
is  C  and  whose  radius  is  CA. 

279,  Generating  a  Sphere.     The  surface 
of  a  sphere  may  be  generated  by  revolv- 
ing a  semicircle  about  its  diameter  as  an 
axis. 

Thus  the  surface  of  the  sphere  CA  may  be 
generated  by  revolving  the  semicircle  MNA 
about  the  diameter  MN. 

280,  COROLLARY  1.     If  two  spheres  have  equal  radii  they  may 
be  made  to  coincide  and  hence  are  equal. 

281,  COROLLARY  2.     All  radii  of  the  same  sphere  or  of  equal 
spheres  are  equal. 

282,  COROLLARY  3.     All  diameters  of  the  same  sphere  or  of 
equal  spheres  are  equal. 

106 


106  SOLID   GEOMETRY:    BOOK    V 

PLANE   SECTION  OF  A  SPHERE 

283,   THEOREM  I.     A  section  of  a  spherical  surface 
made  by  a  plane  is  a  circle. 


Given  a  sphere  with  center  C  cut  by  a  plane  M. 

To  prove  that  the  points  common  to  the  surface  of  the  sphere 
and  the  plane  form  a  circle. 

Proof:  From  the  center  C  draw  CA  perpendicular  to  the 
plane  M. 

Let  B  and  D  be  any  two  points  common  to  the  plane  and 
the  surface  of  the  sphere.  Complete  the  figure  and  prove 
AB  =  AD. 

Hence  any  two  points  common  to  the  surface  of  the  sphere 
and  the  plane  M  are  equidistant  from  the  point  A.  That  is, 
these  points  lie  on  a  circle. 

How  must  this  proof  be  modified  in  case  the  plane  M 
passes  through  the  center  of  the  sphere? 

284,  COROLLARY  1.     Through  three  points  on  a  spherical  sur- 
face there  is  one  and  only  one  circle. 

Suggestion.     How  many  planes  pass  through  these  points  ? 

285.  COROLLARY  2.     A  radius  of  a  sphere  through  the  center 
of  a  circle  on  its  surface  is  perpendicular  to  the  plane  of  the 
circle;  and  conversely,  a  radius  of  a  sphere  perpendicular  to  the 
plane  of  a  circle  on  its  surface  passes  through  the  center  of 
the  circle. 


THE  SPHERE  107 


PROPERTIES   OP   CIRCLES   ON  A  SPHERE 

Figures  on  a  Sphere.     Any  figure  drawn  on  the  surface 
of  a  sphere  is  said  to  lie  on  the  sphere. 

287,  Axis  and  Poles.     The  line  perpen- 
dicular to    the  plane  of  a  circle  at  its 
center  is  called  the  axis  of  the  circle. 

The  points  in  which  the  axis  of  a 
circle  on  a  sphere  meets  the  surface  of 
the  sphere  are  called  the  poles  of  the 
circle. 

288,  Great  and  Small  Circles  on  a  Sphere.     If  the  plane  of  a 
circle  on  a  sphere  passes  through  the  center  of  the  sphere,  it  is 
called  a  great  circle  of  the  sphere,  and  if  not,  it  is  called  a  small 
circle. 

Thus,  in  the  figure,  AB  is  a  small  circle,  PP'  is  its  axis,  and  P  and  P' 
are  its  poles. 

The  circle  passing  through  P  and  P'  and  containing  the  center  C  is  a 
great  circle. 

289,  Inside  and  Outside  of  a  Sphere.     A  point  is  inside,  outside, 
or  on  a  sphere  according  as  its  distance  from  the  center  is  less 
than,  greater  than,  or  equal  to  the  radius  of  the  sphere. 

290,  COROLLARY  1.    The  center  of  a  great  circle  on  a  sphere  is 
the  center  of  the  sphere. 

291,  COROLLARY  2.     AH  great  circles  on  a  sphere  are  equal 
and  any  two  such  circles  bisect  each  other. 

292,  COROLLARY  3.     The  axis  of  any  circle  on  a  sphere  passes 
through  the  center  of  the  sphere. 

293,  COROLLARY  4.     Through  two  given  points  on  a  sphere 
there  is  one  and  only  one  great  circle  unless  these  points  are  at 
opposite  ends  of  a  diameter. 

294,  COROLLARY  5.     Circles  on  a  sphere  formed  by  parallel 
planes  have  the  same  axis  and  the  same  poles. 


108 


SOLID   GEOMETRY:    BOOK  V 


DISTANCES  FROM  POINTS  ON  A  CIRCLE  TO  ITS  POLE 

295,  Distance  on  a  Sphere.     The  spherical  distance,  or  simply 
the  distance  between  two  points  on  a  sphere  is  the  distance 
measured  between  these  points  along  the  minor  arc  of  the  great 
circle  through  them. 

296,  THEOREM  II.    All  points  of  a  circle  on  a  sphere 
are  equidistant  from  either  pole  of  the  circle. 


Given  P  a  pole  of  the  circle  whose  center  is  A,  with  B,  C,  and  D 
any  points  on  this  circle. 

To  prove  that  the  great  circle  arcs  PB,  PC,  and  PD  are  equal. 

Suggestion.     Prove  chord   PB  =  chord  PC  =  chord  PD,  and  hence 
arc  PB  =  arc  PC  =  arc  PD.  §  48 

297,  Polar  Distance  of  a  Circle.     The  spherical  distance  from 
the  points  of  a  circle  on  a  sphere  to  its  nearest  pole  is  called 
the  polar  distance  of  the  circle. 

298,  A  Quadrant.     One  fourth  of  a  great  circle  is  a  quadrant. 

299,  COROLLARY    1.     Tlie  polar    dis- 
tance of  a  great  circle  is  a  quadrant. 

300,  COROLLARY  2.     If  a  point  on  a 
sphere  is  at  a  quadrant's  distance  from 
each  of  two  points  not  at  the  extremities  of 
the  same  diameter,  it  is  the  pole  of  the 
great  circle  through  these  points. 


THE  SPHERE  109 


USE  OF   A   SPHERICAL   BLACKBOARD 

301,  COROLLARY  3.     Polar   distances  of  equal   circles  on  a 
sphere  are  equal. 

302,  COROLLARY  4.    Straight  line-segments  joining  points  of  a 
circle  on  a  sphere  to  one  of  its  poles  are 

equal. 

It  follows  from  the  preceding  theorem  and 
corollaries  that,  if  a  spherical  blackboard  is 
at  hand,  circles  may  be  constructed  on  it  by 
means  of  crayon  and  string  the  same  as  on 
a  plane  blackboard.  Likewise,  curve-legged 
compasses  may  be  used. 

SIGHT   WORK 

1.  If   two  points  are  at  the  extremities  of  the  same  diameter  of   a 
sphere,  how  many  great  circles  can  be  passed  through  these  points  ? 

2.  What  great  circles  on  the  earth's  surface  pass  through  both  poles  ? 
Where  are  the  poles  of  these  circles  located  ? 

3.  If  A  and  B  are  at  opposite  ends  of  a  diameter,  can  a  small  circle  be 
passed  through  them  ? 

4.  If  two  circles  on  a  sphere  have  the  same  poles,  prove  that  their 
planes  are  parallel.     See  §  294. 

5.  What  is  the  locus  of  all  points  on  a  sphere  at  a  quadrant's  dis- 
tance from  a  given  point  ? 

6.  What  is  the  locus  of  all  points  on  a  sphere  at  any  fixed  distance 
from  a  given  point  on  the  sphere  ?     What  is  the  greatest  such  distance 
possible  ?    Discuss  fully. 

7.  If  two  planes  cutting  a  sphere  are  parallel,  what  can  be  said  of  the 
centers  of  the  circles  thus  formed  ?     What  can  be  said  of  the  poles  of  these 
circles  ?    Are  these  statements  true  of  three  or  more  such  circles  ? 

8.  Find  the  locus  of  the  centers  of  a  set  of  circles  on  a  sphere  formed 
by  a  set  of  parallel  planes  cutting  it. 

9.  AB  is  a  fixed  diameter  of  a  sphere.     A  plane  containing  AB  is 
made  to  revolve  about  it  as  an  axis.     Find  the  locus  of  the  poles  of  the 
great  circles  on  the  sphere  made  by  this  revolving  plane.     How  are  the 
points  A  and  B  related  to  this  locus  ? 


110  SOLID  GEOMETRY:    BOOK  V 

CIRCLES  EQUIDISTANT  FROM  THE  CENTER 

303,  THEOREM  III.  If  the  planes  of  two  circles  on 
a  sphere  are  equidistant  from  the  center,  the  circles  are 
equal ;  and  conversely,  if  two  circles  on  a  sphere  are 
equal,  their  planes  are  equidistant  from  the  center. 


FIG.  1  FIG.  2 

Suggestion.     In  the  first  figure  show  that    (1)  if    CA  =  CA',   then 
AB  =  A'B',  and  (2)  if  AB  =  A'B',  then  CA  =  CA'. 

CIRCLES  UNEQUALLY  DISTANT  FROM  THE  CENTER 

304,  THEOREM  IV.     If  the  planes  of  two  circles  on  a 
sphere  are  unequally  distant  from  the  center,  the  circles 
are  unequal,  the  one  nearer  the  center  being  the  greater  ; 
and  conversely,  if  two  circles  on  a  sphere  are  unequal, 
the  plane  of  the  greater  circle  is  nearer  the  center. 

Suggestion.     In  Fig.    2   above,  show   (1)   that    if    CA  <  CA'   then 
AB  >A'B'  •  and  (2)  if  AB  >  A'B'  then  CA  <  CA'. 

305,  Plane  Tangent  to  a   Sphere.     A 
plane  which  meets   a   sphere    in   only 
one  point  is  tangent  to  the  sphere. 

Two  spheres  are  tangent  to  each  other 

if  they  are  both  tangent  to  the  same    /*f 

plane  at  the  same  point. 

A  line,  is  tangent  to  a  sphere  if  it  contains  one  and  only  one 
point  of  the  sphere. 


THE  SPHERE  111 


PLANE  TANGENT  TO  A  SPHERE 

306,  THEOREM  V.  A  plane  tangent  to  a  sphere  is 
perpendicular  to  the  radius  from  the  point  of  tangency  ; 
and  conversely,  a  plane  perpendicular  to  a  radius  at  its 
extremity  is  tangent  to  the  sphere. 


Given  a  sphere  C  with  plane  M  tangent  to  it  at  A. 

To  prove  that  the  radius  CA  is  perpendicular  to  the  plane  M. 

Proof :  (1)  It  is  necessary  to  prove  that  CA  is  perpendicular 
to  every  line  in  M  through  A.  (Why  ?) 

Draw  any  such  line  as  AB. 

The  plane  BAC  cuts  the  sphere  in  a  circle.  Then  AB  is 
tangent  to  this  circle,  and  hence  perpendicular  to  AC. 

Since  CA  is  _L  to  every  line  in  M  through  A,  it  is  _L  to  M. 

(2)  To  prove  the  converse,  note  that  CA  is  the  shortest 
distance  from  C  to  the  plane  M.  §  82 

Hence,  every  point  of  M  except  A  is  exterior  to  the  sphere. 

That  is,  M  is  a  tangent  plane.      (§  305.)  Q.B.  D. 

307,  Sphere  Inscribed  in  a  Polyhedron.     A  sphere  is  said  to 
be  inscribed  in  a  polyhedron  if  every  face  of  the  polyhedron  is 
tangent  to  the  sphere.     The  polyhedron  is  then  said  to  be  cir- 
cumscribed about  the  sphere. 

308,  Polyhedron  Inscribed  in  a  Sphere.     A  polyhedron  is  in- 
scribed in  a  sphere  if  all  its  vertices  lie  in  the  surface  of  the 
sphere.     The  sphere  is  then  circumscribed  about  the  polyhedron. 


112  SOLID  GEOMETRY:    BOOK  V 

SPHERE   INSCRIBED   IN   A   TETRAHEDRON 

309,   PROBLEM.     To  inscribe  a  sphere  in  a  given  tetra- 
hedron. 


D 


Given  a  tetrahedron  D-ABC. 

To  construct  a  sphere  tangent  to  each  of  its  four  faces. 

Construction:  Construct  planes  bisecting  the  dihedral  angles 
whose  edges  are  AB,  BO,  and  CA.  These  three  planes  meet  in 
a  point  P,  which  is  equally  distant  from  the  four  faces  of  the 
tetrahedron.  With  the  point  P  as  a  center  and  a  radius  PE 
equal  to  the  distance  from  P  to  one  of  the  faces  construct  a 
sphere.  This  sphere  is  inscribed  in  the  tetrahedron. 

Proof :  Every  point  in  the  plane  PAB  is  equidistant  from  the 
faces  ABD  and  ABC.  §  123 

Similarly  every  point  in  the  plane  PBC  is  equidistant  from 
the  planes  ABC  and  DBC,  and  every  point  in  the  plane  PAG 
is  equidistant  from  the  planes  ABC  and  DAC. 

Hence,  the  point  P,  common  to  all  three  planes,  is  equidis- 
tant from  each  of  the  four  faces  of  the  tetrahedron. 

.•.  each  plane  is  tangent  to  the  sphere,  and  the  sphere  is 
inscribed  in  the  tetrahedron.  (§  307.)  Q.  B.  F. 

SIGHT  WORK 

Discuss  the  problem  of  finding  a  point  equally  distant  from  three  planes 
two  of  which  are  parallel.  How  many  such  points  are  there  ? 


THE  SPHERE 


113 


SPHERE  CIRCUMSCRIBED  ABOUT  A  TETRAHEDRON 

310,   PROBLEM.     To  find  a  point  equally  distant  from 
the  four  vertices  of  a  tetrahedron. 

p 


Given  the  tetrahedron  P-ABC. 

To  find  a  point  0  equidistant  from  P,  A,  B,  C. 

Construction  :  At  D,  the  middle  point  of  BC,  construct  a  plane 
perpendicular  to  BC. 

This  plane  contains  E,  the  center  of  the  circle  circumscribed 
about  A  PBC  and  also  F,  the  center  of  the  circle  circumscribed 
about  A  ABC.  Why  ? 

In  the  plane  FDE  draw  EG  J.  ED  and  FH  _L  FD. 

Then  EG  and  FH  cannot  be  parallel  and  hence  meet  in  some 
point  O.  Then  O  is  the  point  required. 

Proof :  Since  BC  is  _L  to  the  plane  FDE,  it  follows  that  each 
of  the  planes  PBC  and  ABC  is  J_  to  plane  FDE.  §  117 

Hence  OE  is  _L  plane  PBC  and  OF  is  J_  plane  ABC.     §  114 

Then  0  is  equidistant  from  P,  B,  and  C,  and  also  from  A,  B, 
and  C.  §  81 

Hence  OA=OB=OC=  OP.  Q.  E.  F. 

311.  COROLLARY  1.  A  sphere  may  be  passed  through  any  four 
points,  not  all  of  which  lie  in  the  same  plane. 

312<  COROLLARY  2.  A  sphere  may  be  circumscribed  about  any 
tetrahedron. 


114 


SOLID   GEOMETRY:    BOOK  V 


TO  FIND   THE  DIAMETER  OF   A   SPHERE 

313,  PROBLEM.  To  find  the  diameter  of  a  given 
material  sphere. 

With  any  point  P  of  the  sphere  as  a  pole,  construct  a 
circle,  and  on  this  circle  select  any  three  points  A,  B,  C, 

Using  a  pair  of  compasses,  measure  the  straight  line-seg- 
ments AB,  BO,  CAj  and  construct  the  triangle  AB'C'  equal 
to^LBC. 

Let  B'iy  be  the  radius  of  the  circle  circumscribed  about 
A'B'C'. 

If  PP'  is  the  axis  of  the  circle  ABC  on  the  sphere  and  BD 
the  radius  of  this  circle,  then  BD  =  B'D'. 


Measure  PB  by  means  of  the  compasses. 

Then  PBP'  is  a  right  triangle,  with  BD  perpendicular  to 
its  hypotenuse  PPf. 

PB  and  BD  being  known,  we  may  now  compute  PD  from 
the  right  triangle  PBD,  and  then  compute  PP'  from  the  simi- 
lar triangles  PBD  and  PP'B,  for  the  latter  using  the  relation 
PD:PB  =  PB:PPr  or  PD  X  PP'  =  PB2. 

The  segment  PP'  may  also  be  found  by  geometric  construc- 
tion ;  namely,  by  drawing  a  triangle  equal  to  P'BP. 

Show  how  to  do  this  when  BP  and  BD  are  known. 


THE  SPHERE 


115 


INTERSECTING  SPHERICAL  SURFACES 

314,   THEOREM  VI.    TJie  intersection  of  two  spherical 
surfaces  is  a  circle. 


Proof :  Two  intersecting  spherical  surfaces  may  be  developed 
by  two  intersecting  circles  rotating  about  the  line  connecting 
their  centers  C  and  (7.  Let  A  and  B  be  the  two  points  com- 
mon to  the  two  circles. 

Then  BA  is  _L  CC'. 

(CC'  is  the  JL  bisector  of  BA.)  §  28 

As  the  figure  rotates  about  the  line  (70",  AB  remains  J_  CC' 
and  therefore  all  its  positions  lie  in  a  plane.  §  79 

Also  DB  and  DA  remain  fixed  in  length. 
Therefore  the  points  A  and  B  trace  out  a  circle.  Q.  E.  D. 


SIGHT  WORK 

1.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  a  given  plane 
at  a  given  point. 

2.  Find  the  locus  of  the  centers  of  all  spheres  of  given  radius  tangent 
to  a  given  line  at  a  given  point. 

3.  Find  the  locus  of  the  centers  of  all  spheres  of  given  radius  tangent 
to  a  fixed  plane. 

4.  Find  the  locus  of  the  centers  of  all  spheres  of  given  radius  tangent 
to  a  fixed  line. 

5 .  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  two  given 
intersecting  planes. 


116  SOLID  GEOMETRY:    BOOK  V 

EXERCISES 

1.  The  four  lines  perpendicular  to  the  faces  of  a  tetra- 
hedron at  their  circumcenters  meet  in  a  point. 

2.  The  six  planes  perpendicular  to  the  edges  of  a  tetra- 
hedron at  their  middle  points  all  meet  in  a  point. 

3.  The  planes  bisecting  the  six  dihedral  angles  of  a  tetra- 
hedron all  meet  in  a  point. 

4.  Show  that  a  sphere  may  be  inscribed  in  a  cube. 

5.  Show  that  a  sphere  may  be  circumscribed  about  a  cube. 

6.  Can  a   sphere    be   circumscribed   about  a  rectangular 
parallelepiped  which  is  not  a   cube?     Can  a  sphere  be  in- 
scribed in  it?     Prove. 

7.  If  a  plane  M  is  tangent  to  a  sphere  at  a  point  A9  show 
that  the  plane  of  every  great  circle  of  the  sphere  through  A 
is  perpendicular  to  M. 

8.  Show  that  the  line  of  centers  of  two  intersecting  spheres 
meets  the  spherical  surfaces  in  the  poles  of  their  common  circle. 

9.  Show  that  two  spheres  are  tangent  if  they  meet  on  their 
line  of  centers.     Distinguish  two  cases.     State  and  prove  the 
converse  of  this  proposition. 

10.  If  a  sphere  is  tangent  to  a  given  plane  M  at  a  given 
point  At  how  many  additional   points  on  the  sphere  are  re- 
quired to  determine  it  ? 

Suggestion.     Suppose  one  point  P  given.    Pass  a  plane  through  P  _L 
to  plane  M  at  A.     Is  there  only  one  such  plane  ?    Discuss  fully. 

11.  Describe  the  set  of  all  lines  in  space  whose  distances 
from  the  center  of  a  sphere  are  all  equal  to  the  radius  of  the 
sphere. 

12.  Describe  the  set  of  all  planes  whose  distances  from  the 
center  of  a  sphere  are  all  equal  to  the  radius  of  the  sphere. 

13.  Describe  the  set  of  all  spheres  of  given  radius  which 
are  tangent  to  a  given  sphere  of  greater  radius. 


THE  SPHERE 


117 


SPHERICAL  ANGLES 

315,   Spherical  Angles.     Two  planes  through  the  center  of  a 
sphere  cut  the  surface  of  the  sphere  in  two  great  circles  which 
intersect  in  two  points  and  form  four 
spherical  angles   about   each   of   these 
points.     Any  two  of  these  angles  are 
adjacent  or  vertical  as  in  the  case  of 
angles  formed  by  straight  lines. 

A  spherical  angle  is  measured  by  the 
angle  between  the  tangents  to  its  sides 
(arcs)  at  their  common  point. 

Only  angles  formed  by  great  circles  are  considered  in  this  book. 
MEASURING  A  SPHERICAL  ANGLE 


316,  THEOREM  VII.     A  spherical  angle  is  measured 
by  an  arc  of  the  great  circle  whose  pole  is  the  vertex 
of  the  angle  and  which  is  intercepted  by  the  sides  of 
the  angle. 

Suggestion.     Show  that  AB  measures  the  dihedral  angle  formed  by 
the  planes  PAC  and  PBC  and  that  Z  BCA  =  Z  TPE. 

317,  COROLLARY   1.     The  sum   of  the   consecutive   spherical 
angles  about  a  point  is  four  right  angles. 

318,  COROLLARY  2.     A  spherical  angle  is  equal  to  the  dihe- 
dral angle  formed  by  the  planes  of  its  arcs. 


118 


SOLID  GEOMETRY:    BOOK  V 


SPHERICAL  POLYGONS 


The  portion  of  a 


319,  Polygons  and  Triangles  on  a  Sphere, 
spherical    surface    contained   within  a 
polyhedral  angle  whose  vertex  is  at  the 
center  of  a  sphere  is  called  a  spherical 
polygon. 

It  follows  that  for  every  spherical 
polygon  there  is  a  corresponding  poly-    M 
hedral  angle  at  the  center  of  the  sphere 
made  by  drawing  radii  to  the  vertices 

of  the  polygon.  The  face  angles  of  the  polyhedral  angle  corre- 
spond to  the  sides  of  the  spherical  polygon  and  the  dihedral 
angles  of  the  polyhedral  angle  to  the  angles  of  the  spherical 
polygon. 

Since  a  plane  through  the  center  of  a  sphere  intersects  the 
surface  in  a  great  circle,  it  follows  that  the  sides  of  a  spherical 
polygon  are  arcs  of  great  circles. 

Since  a  plane  may  be  passed  through 
the  vertex  of  a  polyhedral  angle  such 
that  the  polyhedral  angle  lies  entirely 
on  one  side  of  it,  it  follows  that  a 
spherical  polygon  lies  within  one  hemi- 
sphere. 

A  spherical  polygon  of  three  sides  is  a  spherical  triangle. 

320,  Relation  between  the  Parts  of  a  Spherical  Polygon  and 
the  Corresponding  Polyhedral  Angle. 

(1)  The  face  angles  of  the  polyhedral  angle  are  measured  by 
the  arcs  forming  the  sides  of  the  spherical  polygon.     Why  ? 

(2)  The  dihedral  angles  of  the  polyhedral  angle  are  equal 
in  measure  to  the  angles  of  the  spherical  polygon.  §  318 

SIGHT  WORK 

Why  is  the  vertex  of  a  polyhedral  angle  which  is  used  in  defining  a 
spherical  polygon  placed  at  the  center  of  the  sphere  ? 


THE  SPHERE  119 


COROLLARIES  ON  SPHERICAL  POLYGONS 

The  following  propositions   are  now  obvious  corollaries  of 
the  preceding  definitions  and  discussions. 

321,  The,  sum  of  two  sides  of  a  spherical  triangle  is  greater 
than  the  third  side. 

This  is  a  direct  consequence  of  §  151. 

322,  The  sum  of  the  sides  of  a  spherical  polygon  is  less  than 
360°. 

This  is  a  direct  consequence  of  §  152. 

323,  Each  side  of  a  spherical  polygon  is  less  than  180°. 

SIGHT  WORK 

1.  What  is  the  angle  between  a  meridian  on  the  earth's  surface  and 
the  equator  ? 

2.  If  two  meridians  are  drawn  meeting  the  equator  10°  apart,  what 
is  the  angle  between  these  meridians  ? 

3.  If  two  meridians  meet  the  equator  90°  apart,  what  can  be  said  of 
the  three  angles  of  the  spherical  triangle  thus  formed  ? 

4.  If  in  two  spherical  triangles  the  three  sides  of  one  are  equal  re- 
spectively to  the  three  sides  of  the  other,  what  can  be  said  of  the  face 
angles  of  the  corresponding  trihedral  angles  ? 

5.  If  in  two  spherical  triangles  the  three  angles  of  one  are  equal  re- 
spectively to  the  three  angles  of  the  other,  what  parts  are  equal  in  the 
corresponding  trihedral  angles  ? 

6.  If  in  two  spherical  polygons  the  angles  of  one  are  equal  respec- 
tively to  the  angles  of  the  other,  what  parts  are  equal  in  the  correspond- 
ing polyhedral  angles  ? 

7.  If  in  two  spherical  polygons  the  sides  of  one  are  equal  respectively 
to  the  sides  of  the  other,  what  parts  of  the  corresponding  polyhedral 
angles  are  equal  ? 

8.  Prove  that  if  the  sides  of  two  spherical  triangles  are  equal,  then 
the  angles  of  the  triangles  are  equal. 

Suggestion.    Use  §  144. 


120 


SOLID   GEOMETRY:    BOOK   V 


SHORTEST   SPHERICAL   DISTANCE  BETWEEN   TWO  POINTS 

324,  THEOREM  VIII.     The   shortest   distance   on   a 
sphere  between  two  of  its  points   is   measured   along 
the  minor  arc  of  a  great  circle  passing  through  these 
points. 

Proof :  Let  A  and  B  be  any  two  points 
on  a  sphere,  AB  the  minor  arc  of  a  great . 
circle  through  them,  and  ADCB  any  other 
curve  on  the  sphere  connecting  A  and  B. 
Let  D  and  C  be  any  two  points  on  the 
curve  ADCB  taken  in  order  from  A  to  B. 

Draw  the  great  circle  arcs  AD,  AC,  DC,  and  CB.  Then 
by  §  321  AC  +  CB  >  AB  and  AD  +  DC  >  AC. 

Hence,  AD  +  DC  +  CB  >  AB. 

If  in  like  manner  we  subdivide  AD,  DC,  CB,  and  continue 
this  process,  we  obtain  a  succession  of  paths,  each  longer  than 
the  preceding,  and  thus  we  get  closer  and  closer  to  the  length 
of  the  curve  ADCB. 

Hence,  the  curve  ADCB  must  be  greater  than  AB. 

325,  Symmetrical     Spherical     Triangles. 

Two  spherical  triangles  are  symmetrical 
if  the  sides  and  angles  of  one  are  equal 
respectively  to  the  sides  and  angles  of 
the  other,  but  arranged  in  the  opposite 
order.  Compare  §  146. 

SIGHT  WORK 

1.  Is  it  possible  to  move  the  spherical  triangle  ABC  in  the  above 
figure  so  as  to  make  it  coincide  with  triangle  A'B'  C1? 

2.  If  in  two  plane  triangles  AB  C  and  A'B'C'  the  corresponding  parts 
are  equal,  is  it  always  possible  to  make  the  triangles  coincide  ?    Is  this 
always  possible  without  inverting  one  of  the  triangles  ?     Discuss  the  dif- 
ference in  this  respect  between  plane  and  spherical  triangles. 


Q.  E.  D. 


THE  SPHERE 


121 


TRIHEDRAL   ANGLES   AND   SPHERICAL  TRIANGLES 

326,  THEOREM  IX.  Two  equal  or  symmetrical  tri- 
hedral angles  with  vertices  at  the  center  of  a  sphere 
intercept  equal  or  symmetrical  triangles,  respectively, 
on  the  sphere. 


B  A 


Suggestions  for  Proof.  In  case  the  trihedral  angles  are  equal  they 
can  be  made  to  coincide,  whereby  the  spherical  triangles  will  also  be  made 
to  coincide.  The  triangles  are  therefore  equal.  In  case  the  trihedral 
angles  are  symmetrical  it  follows  directly  from  §§  146,  325  that  the  spherical 
triangles  are  symmetrical 

327,  THEOREM    X.     If  the   radii   drawn  from   the 
vertices  of  a  spherical  triangle  are  extended,  they  meet 
the  splwre,  in  the,  vertices  of  a  triangle  symmetrical  to 
the  given  triangle. 

The  proof  is  left  for  the  student. 

328,  THEOREM  XI.      Two  triangles 
on    the   same   sphere,    or    on    equal 
spheres,   are   equal   or    symmetrical 
if  three  sides  of  one  are  equal  respec- 
tively  to    three   sides   of   the   other. 

This  is  a  direct  corollary  of  §§  145,  149.  See  the  figure 
under  §  326. 


122 


SOLID   GEOMETRY:    BOOK  V 


TRIANGLES  EQUAL  OR  SYMMETRICAL 

329,  THEOREM  XII.  Two  triangles  on  the  same 
sphere,  or  on  equal  spheres,  are  equal  or  symmetrical 
if  two  sides  and  the  included  angle  of  one  are  equal 
respectively  to  two  sides  and  the  included  angle  of  the 
other. 


Proof :  This  is  a  direct  corollary  of  the  theorems  §§  142, 150. 

330,  Vertical  Spherical  Angles.     Spherical  angles  are  vertical 
if  they  have  the  same  vertex  and  if  the  sides  of  one  are  exten- 
sions of  the  sides  of  the  other. 

331,  Right   Spherical  Triangle.     A  right   spherical   triangle 
has  one  of  its  angles  a  right  angle. 

332,  Isosceles    Spherical    Triangle.      An    isosceles    spherical 
.  triangle  has  two  equal  sides. 

SIGHT  WORK 

1.  Show  that  vertical  spherical  angles  are  equal. 

2.  What  kind  of  spherical  triangle  is  formed  by  two  meridians  on  the 
earth's  surface  and  the  arc  of  the  equator  which  they  intercept  ? 

3.  Show  that  every  spherical  triangle  formed  as  described  in  Example 
2  contains  more  than  two  right  angles. 

4.  Show  how  to  form  a  spherical  triangle  as  in  Example  2  in  which 
each  angle  shall  be  a  right  angle. 

5.  Show  how  to  form  a  spherical  triangle  as  in  Example  2  in  which 
one  angle  shall  be  179°  and  each  of  the  other  two  90°. 


THE  SPHERE  123 


EQUAL  ANGLES  OPPOSITE  EQUAL  SIDES 

333,  THEOREM  XIII.     The  angles 
opposite  the  equal  sides  of  an  isos- 
celes spherical  triangle  are  equal. 

Suggestion.     Let  AC  and  BC  be  the  equal 
sides.     Draw  CD  to  the  middle  point  of  AB.    A< 
Then  use  §  328. 

334,  COROLLARY.       If     two     isosceles 

spherical  triangles  are  symmetrical,  they  are  equal,  and  conversely. 

EXERCISES 

1.  Compare   fully  the   theorems   on  the  equality  of  plane 
triangles  and  of   trihedral  angles.     Is   there  any  theorem  in 
either  case   for  which   there  is  no  corresponding  theorem  in 
the  other? 

2.  Compare  in  the  same  manner  the  theorems  on  the  equal- 
ity of  plane  triangles  and  of  spherical  triangles. 

3.  Compare  in  the  same  manner  the  theorems  on  the  equal- 
ity of  trihedral  angles  and  of  spherical  triangles. 

4.  If  two  face  angles  of  a  trihedral  angle  are  equal,  the 
opposite  dihedral  angles  are  equal. 

5.  If  two  face  angles  of  a  trihedral  angle  are  equal,  it  is 
equal  to  its  symmetrical  trihedral  angle. 

Suggestion.  Compare  the  sides  and  then  the  angles  of  the  correspond- 
ing spherical  triangles.  Use  §§  333,  329. 

6.  Show  how  to  find  a  pole  of  the   circle   through  three 
given  points  on  a  sphere. 

Suggestion.  Let  the  given  points  be  J.,  5,  C.  By  §  296  a  pole  of 
the  circle  is  equidistant  from  A,  .B,  and  C.  Connect  A  and  B  by  an 
arc  of  a  great  circle  and  construct  another  arc  of  a  great  circle  bisecting 
AB  perpendicularly.  Similarly  construct  a  perpendicular  bisector  of 
BC.  The  points  in  which  these  two  arcs  meet  will  be  the  poles  of  the 
circle  through  J.,  B,  and  C. 


124 


SOLID  GEOMETEY:    BOOK  V 


POLAR  TRIANGLES 

335.  Definition.     If  with,  the  vertices 
of  a  given   spherical   triangle  as   poles 
arcs   of    great    circles    are    constructed, 
another  spherical  triangle  is  formed  which 
is  called  the  polar  triangle  of  the  first. 

Thus  in  the  figure,  A  is  a  pole  of  the  arc 
B'C',  B  is  a  pole  of  the  arc  C'A',  and  C  is  a 
pole  of  the  arc  A'B'.  Hence  A' B'C'  is  the 
polar  triangle  of  triangle  ABC. 

336,  Polar  Triangle.     How   Selected.     If   with   the  vertices 
A,  B,  C  of  a  spherical  triangle  as 

poles  three  complete  great  circles 
are  constructed,  each  of  these 
circles  meets  each  of  the  others 
in  two  points,  thus  forming  eight 
spherical  triangles,  as  shown 
in  the  figure,  namely,  A' B'C', 
A'B'F,  B'C'D,  C'A'E,  A'EF, 
B'DF,  C'DE,  and  DEF. 

There  is  one  and  only  one  of 
these,  namely,  A' B'C',  such  that 
A  and  A  are  on  the  same  side  of  circle  B'C',  B  and  B'  on  the 
same  side  of  circle  AC',  and  C  and  C'  on  the  same  side  of 
circle  A'B'. 

The  triangle  A' B'C'  as  thus  described  is  the  polar  triangle 
of  ABC. 

SIGHT  WORK 

In  the  above  figure  the  parts  of  the  great  circles  which  are  supposed 
to  be  on  the  front  side  of  the  figure  are  given  in  solid  lines  while  the 
parts  on  the  back  side  are  dotted.  Study  the  figure  with  care  and  state 
which  triangles  are  entirely  on  the  front  side,  which  are  entirely  on  the 
back  side  of  the  sphere,  and  which  are  partly  on  the  front  side  and  partly 
on  the  back  side  of  the  sphere. 


THE  SPHERE 


125 


TRIANGLES  POLAR  TO  EACH  OTHER 

337,  THEOREM  XI Y.  If  A'B'C'  is  the  polar  tri- 
angle of  ABCj  then  ABC  is  the  polar  triangle  of 
ARC'. 


Given  A  ABC',  the  polar  triangle  of  ABC. 

To  prove  that  ABC  is  the  polar  triangle  o 

Proof :  (1)  A  is  at  a  quadrant's  distance  from  B  because  B  is 
the  pole  of  AC'.  A  is  also  at  a  quadrant's  distance  from  C 
because  C  is  the  pole  of  A'B'. 

Hence,  A'  is  the  pole  of  BC.  §  300 

Similarly,  B1  is  the  pole  of  AC  and  C'  the  pole  of  AB. 

(2)  To  show  that  A  and  A  lie  on  the  same  side  of  the  cir- 
cle BC,  we  note  that  since  A  is  the  pole  of  the  circle  B'C'  and 
A  lies  on  the  same  side  of  this  circle  with  A',  then  A  and  A 
are  at  less  than  a  quadrant's  distance.  Hence,  it  follows  that 
if  A  is  at  a  quadrant's  distance  from  BC,  A  and  A  must  be 
on  the  same  side  of  BC. 

In  like  manner  we  show  that  B  and  B'  lie  on  the  same  side 
of  AC,  and  C  and  C"  on  the  same  side  of  AB. 

338.  Corresponding  Parts  of  Polar  Triangles.  If  ABC  and 
A'B'C'  are  polar  triangles,  and  if  A  is  a  pole  of  B'C',  then. 
/.  A  and  B'C'  are  said  to  be  corresponding  parts. 


126 


SOLID  GEOMETRY:    BOOK  V 


MEASURE  OF  PARTS  IN  POLAR  TRIANGLES 

339,  THEOREM  XV.  The  sum  of  the  measures  of 
an  angle  of  a  spherical  triangle  and  the  corresponding 
arc  of  its  polar  triangle  is  180°. 

Given  the  polar  triangles  ABC  and  A'B'C'.  Denote  the  meas- 
ures in  degrees  of  the  angles  by  A,  B,  C,  A',  B',  C',  and  of  the 
corresponding  sides  by  a',  b',  c',  a,  b,  c. 

A 


To  prove  that 


A  +  a'  =  180° 
B  +  V  =  180° 


A1  +  a  =  180° 
B1  +  b  =  180° 
C'  +  c  =  180° 

Suggestions  for  Proof :  Extend  (if  necessary)  arcs  A'B'  and 
A'C'  till  they  meet  the  great  circle  BC  in  points  D  and  J£, 
respectively.  Then  arc  DE  is  the  measure  of  Z  ^4'. 

Also  ^  =  90°,  and  DC  =  90°.     (Why  ?) 

But  BE  +  DC=BC+ED  =  a4-A'  =  180°. 


340,  COROLLARY  1.     If  in  two  spherical  triangles  an  angle  of 
one  is  equal  to  an  angle  of  the  other,  then  the  corresponding  sides 
of  their  polar  triangles  are  equal;   and   conversely,  if  in   two 
spherical  triangles  a  side  of  one  is  equal  to  a  side  of  the  other, 
then  the  corresponding  angles  of  their  polar  triangles  are  equal. 

341,  COROLLARY  2.     If  two  spherical  triangles  are  equal  or 
symmetrical,  their  polar  triangles  are  equal  or  symmetrical. 


THE  SPHERE 


127 


SUM  OF  ANGLES  OF  A  SPHERICAL  TRIANGLE 

342,  THEOREM  XVI.  The  sum  of  the  angles  of  a 
spherical  triangle  is  less  than  six  right  angles  and 
greater  than  two  right  angles. 


b' 


Given  the  spherical  triangle  ABC. 
To  prove  that  (1)  Z  A  +  Z  B  +  Z.  C  <  6  rt.  angles. 
(2)  Z  A  +  /.  B  +  /.  C  >  2  rt.  angles. 

Proof:  Construct  the  polar  triangle  A'B'C',  with  sides 
a',  6',  c'. 

(1)  By  §  339  Z  A  +  ^  £  +  ^  C+  a'  +  6'  +  c'  =  6  rt.  A. 
Since    a'  +  6'  +  c'  >  0,  .-.  Z  ^  +  Z  £  +  Z  C  <  6  rt.  A. 

(2)  Using  §  322,  show  that  Z^4-ZJB+ZO>2rt.  A 

343,  COROLLARY  1.     $to£e  cme?  prove  the  theorem  on  trihedral 
angles  which  corresponds  to  Theorem  XVI. 

344,  COROLLARY  2.     The  sum  of  the  angles  of  a  spherical 
polygon  of  n  sides  is  greater  than  2  (n  —  2)  right  angles  and  less 
than  2  n  right  angles. 

Proof :  Divide  the  polygon  into  n  —  2  triangles.  Then  by 
§  342  the  sum  of  the  angles  >  2(n  —  2)  rt.  A. 

Since  each  angle  is  less  than  two  right  angles,  it  follows 
that  the  sum  is  less  than  2  n  right  angles. 

345,  COROLLARY  3.     State  and  prove   the  theorem  on  poly- 
hedral angles  which  corresponds  to  Corollary  2. 


128 


SOLID   GEOMETRY:    BOOK  V 


EQUAL  AND  SYMMETRICAL  SPHERICAL  TRIANGLES 

346,  THEOREM  XVII.     Two  triangles  on  the  same 
sphere,  or  on  equal  spheres,  are  equal  or  symmetrical 
if  two  angles  and  the  included  side  of  one  are  equal 
respectively  to  two  angles  and  the  included  side  of  the 
other. 

Proof.  By  §§  340  and  329  the  polar  triangles  of  the  given 
triangles  are  equal  or  symmetrical.  Hence,  by  §  341,  the 
given  triangles  themselves  are  equal  or  symmetrical. 

347,  COROLLARY.     State  and  prove  the  theorem  on  trihedral 
angles  which  corresponds  to  TJieorem  XVII. 


THEOREM  XVIII.  Two  triangles  on  the  same 
sphere,  or  on  equal  spheres,  are  equal  or  symmetrical 
if  the  angles  of  one  are  equal  respectively  to  the  angles 
of  the  other. 


Proof:  By  §§  340  and  328  the  polar  triangles  of  the  given 
triangles  are  equal  or  symmetrical.  Hence,  by  §  341,  the 
given  triangles  themselves  are  equal  or  symmetrical. 

349,  COROLLARY.  State  and  prove  the  theorem  on  trihedral 
angles  which  corresponds  to  TJieorem  XVIII. 

Is  there  a  theorem  on  plane  triangles  corresponding  to  that 
of  §  348? 


THE  SPHERE 


129 


CONSTRUCTION   OF   SPHERICAL  TRIANGLES 

350,   PROBLEM.     On  a  given  sphere  to  construct  a 
spherical  triangle  when  its  sides  are  given. 


Solution.  Let  0  be  the  given  sphere,  and  a,  b,  c  the  sides  of 
the  required  triangle,  and  let  AA'  be  any  diameter  of  the 
sphere.  With  A  as  a  pole,  construct  circles  DBE  and  FOG, 
whose  polar  distances  from  A  are  c  and  b  respectively. 

With  B  as  a  pole,  construct  a  circle  HCK,  whose  polar  dis- 
tance from  B  is  a.  Then  construct  the  three  great  circle  arcs, 
AB,  BC,  CA.  ABC  is  the  required  triangle.  Why  ? 

SIGHT  WORK 

1.  What  restrictions  if  any  is  it  necessary  to  impose  upon  the  three 
given  sides  of  the  triangle  in  §  350?     (See  §§  321,  322.) 

2.  In  plane  geometry  two  equal  triangles  may  be   constructed  upon 
the  same  base  and  on  the  same  side  of  it.    Is  a  corresponding  construction 
possible  on  the  sphere  ? 

3.  If  in  the  above  construction  each  of  two  sides  of  the  required  triangle 
is  very  great,  that  is,  nearly  a  semicircle,  show  from  the  construction  that 
the^third  side  must  be  very  small. 


130  SOLID  GEOMETRY:    BOOK  V 

CONSTRUCTION  OP  SPHERICAL  TRIANGLES 

351,  PROBLEM.     To   construct   a   spherical   triangle 
when  its  three  angles  are  given. 

Solution.  Let  the  three  given  angles  be  A,  B,  C,  and  let  af, 
&',  c'  be  arcs  such  that  a'  +  Z  A  =  180°,  &'  +  Z  B  =  180°,  and  c'+ 
Z  (7  =  180°.  Then  the  triangle  whose  arcs  are  a',  6',  c'  will  be 
the  polar  triangle  of  the  required  triangle.  This  latter  triangle 
A'B'C'  may  be  constructed  by  the  method  of  §  350.  Then  con- 
struct the  polar  triangle  of  A'B'C',  which  will  be  the  required 
triangle. 

Give  reasons  in  full  for  each  step. 

352,  PROBLEM.     To  construct  a  trihedral  angle  when 
its  face  angles  are  given. 

Solution.  Construct  the  corresponding  spherical  triangle  by 
the  method  of  §  350. 

Give  the  construction  in  full  and  prove  each  step. 

353,  PROBLEM.     To  construct  a  trihedral  angle  when 
its  dihedral  angles  are  given. 

Solution.  Construct  the  corresponding  spherical  triangle  by 
the  method  of  §  351. 

Give  reasons  in  full  for  each  step. 

SIGHT  WORK 

1.  If  two  spherical  triangles  having  angles  respectively  equal  are  con- 
structed on  the  same  sphere,  how  are  these  triangles  related  ?    Prove. 

2.  If  two  trihedral  angles  with  face  angles  respectively  equal  are  con- 
structed as  in  §  352,  how  are  they  related  ?    Prove. 

3.  If  two  trihedral  angles  with  dihedral  angles  respectively  equal  are 
constructed  as  in  §  353,  how  are  the  trihedral  angles  related  ?    Prove. 

4.  What  restrictions  if  any  must  be  placed  upon  the  given  angles 
J.,  5,  C  in  §  351?    Compare  Example  1,  page  129. 

5.  What  restrictions  if  any  are  needed  in  Examples  2  and  3  ? 


THE  SPHERE  131 


EXERCISES 

1.  If  two  angles  of  a  spherical  tri- 
angle are  unequal,   the   sides   opposite 
them  are  unequal,  the  greater  side  being 
opposite  the  greater  angle. 

Suggestion.     In  the  triangle  A B  C  let  Z.  B  be  greater  than  Z  A.     Draw 
BD,  making  Z  ABD  =-.  Z  A. 

Then,  AI>  =  BD,  and  BD  +  Dd  >  BC. 

Hence,  show  that  AC  >  BC. 

2.  State   and   prove   a   theorem   on   trihedral   angles  cor- 
responding to  the  preceding. 

3.  If  the  sides  of  a  spherical  triangle  are  60°,  80°,  120°,  find 
the  angles  of  its  polar  triangle. 

4.  If  the  angles  of  a  spherical  triangle  are  72°,  104°,  88°, 
find  the  sides  of  the  polar  triangle. 

5.  If  a  triangle  is  isosceles,  prove  that  its  polar  triangle  is 
isosceles. 

6.  If  each  side  of  a  spherical  triangle  is  a  quadrant,  describe 
its  polar  triangle. 

7.  Is  it  possible  to  construct  a  spherical  triangle  whose 
sides  are  50°,  60°,  120°? 

8.  Is  it  possible  to  construct  a  spherical  triangle  whose 
sides  are  100°,  120°,  150°? 

Suggestion.     Consider  the  polar  triangle  of  such  triangle. 

9.  Consider  the  questions  on  trihedral  angles  corresponding 
to  the  two  preceding. 

10.  If  the  sides  of  a  spherical  triangle  are  75°,  95°,  and  115° 
respectively,  find  the  angles  of  each  triangle  formed  by  the 
polar  construction. 

11.  If  it  is  given  that  a  spherical  triangle  is  equilateral,  can 
we  infer  from  the  theorems  thus  far  proved  that  its  polar 
triangle  is  equilateral  ? 


132  SOLID  GEOMETRY:    BOOK  V 

SYMMETRICAL  TRIANGLES  ARE  EQUAL  IN   AREA 

354,  Equal  Areas  Defined.     Two  spherical  polygons  are  said  to 
have  equal  areas  if  they  can  be  made  to  coincide,  or  if  they  can 
be  divided  into  parts  which  can  be  made  to  coincide  in  pairs. 

Compare  this  with  the  definition  of  equal  areas  in  Plane  Geometry. 

355.  THEOREM    XIX.     Two   symmetrical   spherical 
triangles  are  equal  in  area. 


Proof :  Let  ABC  be  one  of  the  given  triangles.  Extend  the 
radii  AO,  BO,  CO  to  meet  the  sphere  in  Alt  B^  Ci,  thus  form- 
ing a  triangle  symmetrical  to  A  ABC.  §  327 
Let  P  be  a  pole  of  the  circle  through  A,  B,  C.  Extend  PO 
to  meet  the  sphere  in  Plt  Draw  PA,  PB,  PC,  and  J\Alt 
and  PjOi. 

Suppose  that  P  lies  within  A  ABC. 
The  spherical  triangles  PAB,  PBC,  PCA,  P^B 
PiQAi  are  all  isosceles.  §  296 

Now  prove 

(1)  A  PAB  =  A  P.A.B, ;         (2)  A  PBC  =  A  P&d  ; 

(3)  APCA  =  AP1C1A1. 
.-.  area  A  ABC  =  area  A  AiB^d* 

But  A  AiBiCi  is  equal  to  any  other  triangle  which  is  sym- 
metrical to  A  ABC. 

.-.  Two  symmetrical  spherical  triangles  are  equal  in  area. 


THE  SPHERE 


133 


BIRECTANGULAR  SPHERICAL  TRIANGLES 

356,   A  birectangular  spherical  triangle  is  one  having  two  right 
angles,  as  Z  PAB  in  the  figure. 

If  the  third  angle  of  a  birectangular  triangle 
is  1°,  the  triangle  contains  one  of  720  equal 
parts  of  the  surface  of  the  sphere. 


357,  Spherical  Degree.     The  area  of  a  birec- 
tangular triangle  having  a  vertex  angle  of  one 
degree  is  called  a  spherical  degree  and  is  used 
as  a  unit  of  measure  of  areas  of  spherical  poly- 
gons. 

In  a  similar  manner  we  define  a  spherical  minute  and  a 
spherical  second. 

358,  The  Lune.     A  lune  is  a  figure  formed 
by  two  great  semicircles  having  the  same  end- 
points.     The  angle  between  these  semicircles 
is  the  angle  of  the  lune. 

359,  COROLLARY  1.     The  area  of  a  birectan- 
gular spherical  triangle  in  terms  of  spherical  de- 
grees is  equal  to  the  number  of  degrees  in  the 
third  angle  of  the  triangle. 

360,  COROLLARY  2.     The  area  of  a  lune  in  terms  of  spherical 
degrees  is  twice  the  number  of  degrees  in  the  angle  of  the  lune. 

361,  Spherical  Excess.     The  number  of  degrees  by  which 
the  sum  of  the  angles  of  a  spherical  triangle  exceeds  180°  is 
called  the  spherical  excess  of  the  triangle. 

The  spherical  excess  of  a  spherical  polygon  is  the  sum  of  its 
angles  less  (n  —  2)180°,  where  n  is  the  number  of  sides  of  the 
polygon. 

EXERCISE 

Show  that  the  spherical  excess  of  a  spherical  polygon  is  less 
than  four  right  angles. 


134  SOLID  GEOMETRY:    BOOK  V 

AREA  OF  A  SPHERICAL  TRIANGLE 

362,  THEOREM  XX.  The  area  of  a  spherical  triangle 
in  terms  of  spherical  degrees  is  equal  to  its  spherical 
excess.  * 


Proof  :  We  are  to  show  that  area  A  ABC  = 
-  180°. 

By  §  360,  the  areas  of  the  lunes  ACDB,  CAEB,  BCFA  in 
spherical  degrees  are  as  follows  : 

A  ABC  +  A  BCD  =  ACDB  =  2  Z  A. 
AABC+ABAE=CAEB  =  2^  C. 
A  A  B  C  +  A  OF  A  =  B  CFA  =  2  Z  £. 
Hence,  adding, 

3A  ABC+  &BCD,  BAE,  CFA  =  2(Z.  A  +  Z.  B  +  ^  C). 
Now  A  BCD  and  AEF  are  symmetrical  and  equal  in  area. 
Hence, 

2  A  ABC  +  A  ABC,  AEF,  BAE,  CFA  =  2UA  +  ^B  +  ^O). 
But  A  ABC,  AEF,  BAE,  CFA  together  constitute  a  hemi- 
sphere or  360  spherical  degrees. 
Hence,  2  A  ABC  +  360°  =  2(Z^  +  Z5  +  ZC). 
Solving,  &ABC  =  ZA  +  Z  B  +  ZO-1800.    Q.E.D. 


NOTE.  —  The  spherical  degree  differs  fundamentally  from  the  units 
of  measure  hitherto  used.  This  unit  is  a  certain  fraction  of  the  sur- 
face of  the  sphere  and  hence  its  actual  size  depends  upon  the  size  of  the 
sphere. 

*  This  theorem  was  discovered  by  Cavalieri.     See  Frontispiece. 


THE  SPHERE  135 


AREA  OF  A  SPHERICAL  POLYGON 

363,  THEOREM  XXI.  The  area  of  a  spherical  poly- 
gon in  terms  of  spherical  degrees  is  equal  to  its  spheri- 
cal excess. 

Proof:  Join  one  vertex  of  the  polygon  to  each  non-adjacent 
vertex,  thus  forming  n  —  2  spherical  triangles.  Now  since  the 
sum  of  the  spherical  excesses  of  these  triangles  is  the  spheri- 
cal excess  of  the  polygon,  the  conclusion  is  evident. 

SIGHT  WORK 

1.  What  is  the  area  in  spherical  degrees  of  a  birectangular  triangle 
if  one  of  its  angles  is  54°  ?  if  one  angle  is  24°  ?  if  one  angle  is  36°  ? 

2.  What  is  the  spherical  excess  of  a  triangle  the  sum  of  whose  angles 
is  285°  ? 

3.  What  is  the  spherical  excess  of  a  triangle  whose  angles  are  75°, 
110°,  and  150°? 

•  4.  What  is  the  spherical  excess  of  a  spherical  hexagon  the  sum  of 
whose  angles  is  1060°  ? 

5.  What  is  the  spherical  excess  of  a  spherical  pentagon  whose  angles 
are  90°,  120°,  75°,  108°,  and  165°  ? 

6.  What  is  the  angle  of  a  lune  whose  area  is  160  spherical  degrees. 

7.  The  spherical  excess  of  a  triangle  is  120°.     Two  of  its  angles  are 
110°  and  108°  respectively.     Find  the  third  angle. 

8.  Between  what  limits  is  the  sum  of  the  angles  of  a  spherical  poly- 
gon of  eight  sides  ? 

9.  If  the  sum  of  the  angles  of  a  spherical  polygon  is  11  right  angles, 
what  is  known  about  the  number  of  its  sides  ? 

10.  If  the  sum  of  the  angles  of  a  spherical  polygon  is  14  right  angles, 
what  is  known  about  the  number  of  its  sides  ? 

11.  The  sides  of  a  spherical  triangle  are  85°,  95°,  110°.     Find  the 
area  of  each  of  the  eight  triangles  formed  by  the  polar  construction  from 
this  triangle. 

12.  The  area  of 'a  spherical  triangle  is  74  spherical  degrees.     One 
angle  is  105°.     Of  the  other  two  angles  one  is  twice  the  other.    Find  all 
the  angles  of  the  triangle. 


136 


/SOLID  GEOMETRY:    BOOK  V 


LATERAL  AREA  OF  A  FRUSTUM  OF  A  CONE 

364,  THEOREM  XXII.  The  lateral  area  of  a  frus- 
tum of  a  right  circular  cone  is  equal  to  the  altitude  of 
the  frustum  multiplied  by  the  circumference  of  a  circle 
whose  radius  is  the  perpendicular  distance  from  a  point 
in  the  axis  of  the  frustum  to  the  middle  point  of  an 
element. 


Given  a  frustum  with  an  element  AA'  whose  middle  point  is  B, 
CC  the  axis  of  the  frustum,  BD  J_  CC  and  EB  JL  AA1. 

To  prove  that  the  lateral  area  is  equal  to  2  TT  •  EB  -  CC'. 
Proof :  By  §  269,  the  lateral  area  is  2V  •  BD  -  AA1. 
Hence,  we  must  show  that 

EB  •  CO  =  BD  -  AA1. 

To  do  this,  draw  A'F  _L  AC  and  show  that 
A  AFA'  ~  A  EDB. 

365,  COROLLARY.  The  lateral  area  of  a  right 
circular  cone  is  equal  to  its  altitude  times  the  length 
of  a  circle  whose  radius  is  the  perpendicular  from 
a  point  in  the  axis  to  the  middle  point  of  an  element. 

Note.  The  above  theorem  and  corollary  are  needed  in  deducing  the 
area  of  the  surface  of  a  sphere.  We  have  already  computed  the  area  of  a 
spherical  triangle  in  terms  of  spherical  degrees,  but  we  now  wish  to  de- 
rive the  area  of  the  spherical  surface  in  terms  of  plane  units  of  measure. 


THE  SPHERE  137 


CIRCUMSCRIBED  AND  INSCRIBED  CONES  AND  FRUSTUMS 

366,  About  a  circle  circumscribe  a  polygon  as  follows  :  Con- 
struct two  diameters  AB  and  CD  at  right  angles  to  each  other 
and  divide  each  quadrant  into  an  even  number  of  parts  by 
points,  as  E,  F,  G.     At  each  alternate 

division  point,  beginning  with  the  first 
point  E,  draw  a  tangent. 

There  results  a  regular  polygon  with 
two  vertices  on  each  of  the  diameters 
AB  and  CD  extended. 

If  now  we  construct  another  polygon 
in  the  same  manner  by  dividing  each 
quadrant  into  twice  as  many  arcs,  it  will 
likewise  have  two  vertices  on  each  of 
the  diameters  AB  and  CD  extended.  The  vertices  on  AB  will 
lie  between  M  and  N. 

This  construction  may  be  repeated  at  pleasure,  thus  obtain- 
ing a  sequence  of  polygons  each  lying  closer  to  the  circle  than 
the  preceding. 

'Now  inscribe  a  polygon  similar  to  the  first  one  of  the  set  just 
circumscribed,  by  joining  the  points  Oand  F,  F  and  A,  and  so  on, 
repeating  this  process  to  form  a  sequence  of  inscribed  polygons. 

If  now  the  whole  figure  is  made  to  revolve  about  AB  as  an 
axis,  the  circle  generates  a  sphere,  and  the  circumscribed  and 
inscribed  polygons  generate  sets  of  circumscribed  and  inscribed 
cones  and  frustums  of  cones. 

367,  Fundamental  Assumption  on  the  Area  of  a  Sphere.     We 

assume  that 

A  sphere  has  a  definite  area  which  is  less  than  the  surface  of 
any  circumscribed  figure  and  greater  than  the  surface  of  any 
inscribed  convex  figure. 

The  student  should  note  that  while  the  statement  just  preceding  is 
obviously  true,  it  is  not  capable  of  proof  by  pure  deduction.  It  is  there- 
fore necessarily  in  the  nature  of  an  assumption  or  axiom. 


138 


SOLID   GEOMETRY:    BOOK  V 


FORMULA  FOR  THE  AREA  OF  A  SPHERE 

368,  THEOREM  XXIII.     The  area  of  a  sphere  whose 
radius  is  r  is  4  Trr2. 

Proof :  (a)  Denote  by  r  the  radius  of  the  circle  which  gener- 
ates the  sphere.     That  is,  in  the  figure,  r  —  OG  =  OE  =  OH. 


'H 


By  §  364  the  lateral  areas  of  the  outside  frustums  whose  axes 
are  OL  and  OK  are  2  Trr  •  0£  and  2  -n-r  •  OK,  and  by  §.  365  the 
lateral  areas  of  the  cones  whose  axes  are  LN  and  KM  are  re- 
spectively 2  Trr  •  LN  and  2  irr  •  KM. 

Hence,  the  total  surface  of  the  whole  circumscribed  figure  is 


2  Trr .  M. K  +  2  Trr  •  K 0  +  2  Trr  •  OL  +  2  wr  •  LN, 
or  2  TTT  (JfZf  +  KO  +OL  +  LN)  =2  Trr  •  Jf2V. 

If  now  a  polygon  of  twice  the  number  of  sides  is  con- 
structed, as  described  in  §  366,  we  obtain  another  circumscribed 
figure  whose  area  is  2  Trr  •  M'N'  where  M '  and  N'  are  the  two 
vertices  on  the  line  AB  extended,  but  lying  between  M  and  N, 
so  that  2  Trr  •  M'N1  <  2  Trr  •  MN. 

As  this  process  goes  on,  the  total  surface  generated  decreases 
and  may  be  made  to  approximate  as  nearly  as  we  please  to 

2  TTT  x  A  B  =  2  Trr  x  2  r  =  4  Trr2. 
Therefore,  the  area  of  the  sphere  cannot  be  greater  than 


THE  SPHERE  139 


(6)  Using  the  same  figure,  let  OG'  be  the  apothem  of  the 
inscribed  polygon. 

Then,  as  under  (a),  we  find  that  the  area  of  the  figure  de- 
veloped by  revolving  the  inscribed  polygon  about  AB  is 

27T-AB-  OG'  =  ±7rr.  OG'. 


By  continuing  to  double  the  number  of  sides,  OG'  may  be  made 
to  approach  as  nearly  as  we  please  to  OG  =  r,  OG'  being 
always  less  than  OG. 

Hence,  the  total  surface  developed  increases  and  approaches 
as  nearly  as  we  please  to 

47rr  x  OG  =  4:7rrz. 

Therefore  the  area  of  the  sphere  cannot  be  less  than  4  Trr2. 

Since  from  (a)  and  (b)  the  area  of  the  'surface  of  the  sphere 
can  be  neither  greater  than  4  Trr2  nor  less  than  4  Trr2,  it  follows 
that  it  is  exactly  equal  to  4  Trr2. 

Q.  B.D. 


SIGHT  WORK 

Use  the  value,  ir  =  3.1416. 

1.  Find  the  surface  of  a  sphere  whose  radius  is  one  inch. 

2.  Find  the  surface  of  a  sphere  whose  radius  is  10  inches. 

3.  What  is  the  relation  between  the  surface  of  a  sphere  and  the  area 
of  a  circle  of  the  same  radius  ? 

4.  Find  the  area  in  square  inches  of  a  spherical  degree  on  a  sphere  of 
radius  15  inches. 

5.  What  fraction  of  the  surface  of  a  sphere  is  occupied  by  a  birec- 
tangular  triangle  having  one  angle  of  65°  ?  by  a  lune  with  an  angle  of  84°  ? 

6.  Find  the  area  in  square  inches  of  a  birectangular  spherical  triangle 
with  one  angle  equal  to  35°,  if  the  triangle  is  on  a  sphere  of  radius 
10  inches. 

7.  Find  the  area  in  square  inches  of  a  lune  whose  angle  is  42°,  if  the 
lune  is  on  a  sphere  of  radius  20  inches. 


140  SOLID   GEOMETRY:    BOOK  V 

EXERCISES  IN  ARITHMETIC  COMPUTATION 

Example.  Find  the  area  in  square  inches  of  a  spherical 
triangle  whose  angles  are  80°,  85°,  and  97°  if  the  radius  of  the 
sphere  is  6  inches. 

Solution.     The  spherical  excess  of  the  triangle  is 

80°  +  85°  +  97°  -  180°  =  262°  -  180°  =  82°. 

The  total  area  of  the  sphere  is  4?r  x  62  =  452.3904  sq.  in.,  and  one  spheri- 
cal degree  is  7^  of  452.3904  =  .62832  sq.  in. 

Hence  82  spherical  degrees  is  82  x  .62832  =  51.52224  sq.  in. 

1.  Find  the  area  in  square  inches  of  a   spherical  triangle 
whose  angles  are  70°,  80°,  90°  if  the  radius  of  the  sphere  is 
10  inches. 

Suggestion.     First  find  the  spherical  excess  of  the  triangle. 

2.  Find  the  area  in  square  inches  of  a  spherical   triangle 
whose  angles  are  95°,  110°,  75°  if  the  radius  of  the  sphere  is 
15  inches. 

3.  Find  the  area  of  a  spherical  polygon  whose  angles  are 
110°,  120°,  130°,  and  95°  if  the  radius  of  the  sphere  is  8  inches. 

4.  Find  the  area  of  a  spherical  polygon  whose  angles  are 
130°,  140°,  110°,  100°,  160°,  and  150°  if  the  radius  of  the  sphere 
is  20  inches. 

EXERCISES  IN  ALGEBRAIC  COMPUTATION 

1.  The  area  of  a  sphere  is  s  square  inches.     Find  the  radius 
of  the  sphere  in  terms  of  s. 

2.  The  spherical  excess  of  a  spherical  triangle  is  e  and  the 
radius  of  the  sphere  is  r.     Find  the  area  of  the  triangle  in 
terms  of  e  and  r. 

3.  The  spherical  excess  of  a  spherical  polygon  is  e  and  its 
area  is  a  square  inches.     Find  the  radius  of   the   sphere   in 
terms  of  e  and  a. 

4.  The  area  in  square  inches  of  a  spherical  triangle  is  a  and 
the  radius  of  the  sphere  is  r.     Find  the  spherical  excess  in 
terms  of  a  and  r. 


THE  SPHERE  141 


THE  VOLUME  OP  A  SPHERE  BY  INSPECTION 

369,  The  formula  for  the  volume  of  a  sphere  may  be  in- 
ferred directly  from  the  accompanying  figure. 

The  sphere  is  covered  with  a  network 
of  spherical  quadrilaterals.  If  these  are 
taken  small  enough,  they  may  be  regarded 
as  approximately  plane  surfaces. 

On  this  supposition  we  have  a  set  of 
pyramids  with  a  common  altitude  r  and 
the  sum  of  their  bases  approximately 
equal  to  the  area  of  the  sphere. 

Hence,  their  combined  volume  is  1  r  x  (area  of  sphere) 
or  1  r  •  4  Ti-r2.  That  is,  the  volume  is  f  irr3. 

It  is  clear  that,  by  making  these  quadrilaterals  sufficiently 
small  each  one  may  be  made  to  approach  as  nearly  as  we 
please  to  a  plane  surface.  Hence,  each  pyramidal  figure  with 
vertex  at  0  is  made  to  approach  the  form  of  a  true  pyramid 
as  nearly  as  we  please. 

370.  Fundamental  Assumption  on  the  Volume  and  Surface  of  a 
Sphere.     In  the  formal  proof  on  page  142,  we  use  polyhedrons 
circumscribed  about  the  sphere  and  we  assume  that 

The  surface  and  the  volume  of  a  sphere  may  be  approximated 
as  nearly  as  we  please  by  taking  the  surfaces  and  the  volumes 
of  a  series  of  circumscribed  polyhedrons  all  of  whose  faces  are 
made  to  decrease  indefinitely. 

Note.  This  assumption,  when  taken  together  with  §  367,  implies  that 
the  surface  obtained  by  taking  the  surfaces  of  a  series  of  polyhedrons 
is  the  same  as  the  surface  obtained  by  taking  the  surfaces  of  the  circum- 
scribed figures  described  in  §  366.  This  is  of  course  obvious  at  a  glance, 
though  a  formal  deductive  proof  is  very  difficult. 

EXERCISE 

Using  §  369  find  the  volume  of  a  sphere  whose  radius  is 
10  in. ;  also  one  whose  diameter  is  6  ft. 


142 


SOLID   GEOMETRY:    BOOK   V 


FORMULA  FOR   THE  VOLUME  OF   A  SPHERE 

371.  THEOKEM    XXIV.       The    volume  of  a   sphere 
whose  radius  is  r  is  -f  Trr3. 

Proof:  Consider  a  sphere  of  radius  r  with  a  cube  circum- 
scribed about  it.     Connect  its  vertices  with  the  center  of  the 


sphere.  Then  the  cube  is  divided  into  six  pyramids,  each  hav- 
ing an  altitude  r.  Since  the  volume  of  each  pyramid  equals 
the  area  of  its  base  times  one  third  its  altitude,  it  follows  that 
the  sum  of  the  volumes  of  these  pyramids  is  equal  to  the  sum 
of  their  bases  times  one  third  their  common  altitude. 

Consider  now  a  polyhedron  obtained  from  this  cube  by  cut- 
ting off  its  eight  vertices  by  planes  tangent  to  the  sphere. 
This  new  polyhedron  is  circumscribed  about  the  sphere  and 
approaches  it  more  nearly  than  the  cube.  Continuing  in  this 
manner  we  may  obtain  a  series  of  circumscribed  polyhedrons 
approaching  the  sphere  as  nearly  as  we  please. 

By  joining  the  vertices  of  each  polyhedron  to  the  center  of 
the  sphere  we  obtain  a  set  of  pyramids,  each  with  altitude  r. 

Suppose  the  total  surfaces  of  these  successive  polyhedrons 
are  s^  s2>  ss>  $4?  •••  •  Then  their  volumes  are  -gSj,  ^s2,  fs3,  •••• 
But,  by  §  370,  slt  s2,  sS)  •••  approach  the  surface  of  the  sphere, 
or  4  Trr2,  as  nearly  as  we  please.  Hence  the  successive  volumes 
approach  -|  x  4  Trr2  =  l^rr3  as  nearly  as  we  please.  That  is, 

Surface  of  a  sphere  =  4  irr2,  and  Volume  of  a  sphere  = 


THE  SPHERE  143 


SIGHT  WORK 

1.  Given  a  sphere  of  radius  6  inches,  is  there  any  upper  limit  to  the 
volume  of  its  circumscribed  polyhedrons  ?    That  is,  can  polyhedrons  be 
circumscribed  having  a  volume  as  large  as  we  please  ? 

2 .  With  the  same  sphere  is  there  any  lower  limit  to  the  volume  of  its 
circumscribed  polyhedrons  ? 

3.  Show  that  the  areas  of  two  spheres  are  in  the  same  ratio  as  the 
squares  of  their  radii  or  of  their  diameters. 

4.  Show  that  the  volumes  of  two  spheres  are  in  the  same  ratio  as  the 
cubes  of  their  radii  or  of  their  diameters. 

5.  Express  the  area  of  a  sphere  whose  radius  is  8  in.  in  terms  of  TT. 

6.  Express  the  volume  of  a  sphere  whose  radius  is  10  ft.  in  terms 
of  TT. 

7.  The  surface   of   a  polyhedron   circumscribed  about  a  sphere   of 
radius  4  in.  is  420  sq.  in.     Find  its  volume. 

EXERCISES   IN   ARITHMETIC   COMPUTATION 

1.  The   volume   of    a   polyhedron    circumscribed    about  a 
sphere  of  radius  3.5  in.  is  450  cu.  in.     Find  its  surface. 

2.  If  the  area  of  a  sphere  is  227  sq.  ft.,  find  its  radius. 

3.  If  the  volume  of  a  sphere  is  335  cu.  in.,  find  its  radius. 

4.  If  the  volumes  of  two  spheres  are  27  cu.  in.  and  729 
cu.  in.,  find  the  ratio  of  their  radii. 

EXERCISES   IN   ALGEBRAIC   COMPUTATION 

1.  The   volume   of  a   sphere  is    v   cubic   inches.     Find  its 
radius  in  terms  of  v  and  it. 

2.  The  area  of  the  surface  of  a  sphere  is  s  square  inches. 
Find  its  volume  in  terms  of  s  and  TT. 

3.  The  area  of  the  surface  of  a  sphere  is  numerically  equal 
to  its  volume.     Find  the  radius  of  the  sphere  if  an  inch  is 
the  unit  of  measure. 

4.  The  difference  between  the  volume  of  a  cube  and  that 
of  its  inscribed  sphere  is  v.     Find  the  radius  of  the  sphere  in 
terms  of  v  and  IT. 


144 


SOLID   GEOMETRY:    BOOK   V 


SPHERICAL  SEGMENTS,   CONES,   SECTORS 

372,  Zone,  Segment.     That   part  of  a  spherical  surface  in- 
cluded between   two   parallel   planes    cutting   it  is  called  a 
zone.     The  perpendicular  distance  between  the  planes  is  the 
altitude  of  the  zone  and  of  the  corresponding  segment. 

The  portion  of  a  sphere  included  between  two  parallel 
planes  cutting  it  is  called  a  spherical  segment,  and  the  two 
circular  sections  made  by  the  parallel  planes  are  its  bases. 

If  one  of  the  cutting  planes  is  tangent  to  the  sphere,  then 
the  spherical  segment  and  the  corresponding  zone  are  said  to 
have  but  one  base.  The  altitude  in  this  case  is  the  perpendic- 
ular distance  from  the  base  to  the  tangent  plane. 

373,  Spherical  Cone,  Sector.     If  one  nappe 
of  a  convex  conical  surface  has  its  vertex 
at  the  center  of  a  sphere,  the  portion   of 
the  sphere  cut  out  by  this  surface  is  called 
a  spherical  cone. 

If  two  spherical  cones  have  the  same  axis, 
one  lying  within  the  other,  the  figure  formed 
by  their  two  lateral  surfaces,  together  with 
the  part  of  the  sphere  intercepted  between 
them,  is  called  a  spherical  sector. 

If  the  two  cones  are  right  circular  cones,  they 
intercept  circles  on  the  sphere,  and  the  zone  thus 
included  is  called  the  base  of  the  spherical  sector. 

If  the  accompanying  figure  be  revolved  about  LM  as 
an  axis,  then  any  arc,  as  GD  or  MD,  generates  a  zone, 
the  former  with  two  bases,  the  latter  with  one. 

The  figure  MDF  or  FDGE  generates  a  spherical  seg- 
ment, the  former  with  one  base,  the  latter  with  two. 

The  figure  CAL  or  CBL  generates  a  spherical  cone, 
CL  being  the  common  axis. 

The  figure  CBA  generates  a  spherical  sector  and  arc  AB  generates  the 
zone  which  is  the  base  of  the  spherical  sector. 


THE  SPHERE  145 


AREA  OF  A  ZONE.   VOLUME  OF  A  SPHERICAL  CONE 

374,   Area  of  a  Zone.     An  argument   precisely  like  that  of 
§  368  shows  that  the  area  of  a  zone  is 


where  h  is  the  altitude  of  the  zone. 

That  is,  instead  of  AB,  the  diameter  in  case  of  the  sphere, 
we  should  have  the  sum  of  the  altitudes  of  the  frustums  cir- 
cumscribed about  the  zone  equal  to  h,  the  altitude  of  the  zone. 

375.   Volume  of  a  Spherical  Cone  and  a  Spherical  Sector.     An 

argument  precisely  like  that  of  §  371  shows  that  the  volume 
of  a  spherical  cone  is 


where  s  is  the  area  of  the  zone  of  one  base  which  is  cut  out  of 
the  sphere  by  the  cone.  Hence,  if  h  is  the  altitude  of  this 
zone,  we  have 


In  like  manner  the  volume  of  a  spherical  sector  is 


where  h  is  the  altitude  of  the  zone  of  two  bases  which  is  cut 
out  by  the  sector. 

EXERCISES 

1.  The  radius  of  a  sphere  is  6  in.  and  the  altitude  of  a  zone 
is  5  in.     Find  the  area  of  the  sphere  and  of  the  zone. 

2.  The  area  of  a  zone  is  36  ir  sq.  ft.  and  its  altitude  4  ft. 
Find  the  radius  of  the  sphere. 

3.  On  a  sphere  of  radius  8  in.  a  spherical  cone  cuts  out  a 
zone  whose  altitude  is  2  in.    Find  the  volume  of  the  cone. 

4.  Find  the  volume  of  the  spherical  sector  cut   out  of  a 
sphere  of  radius  9  in.,  if  the  altitude  of  the  zone  is  2  in. 


146  SOLID  GEOMETRY:    BOOK  V 

VOLUME  OF  A  SPHERICAL   SEGMENT 

376,  PROBLEM.     To  find  the  volume  of  a  spherical 
segment. 

Solution.     Let  r  be   the   radius   of   the        / 
sphere,   and   r^  and   rz   the   radii   of    the      /— 
bases  of  the  segment,  h   the   altitude   of 
the  segment,  and  let  the  segment  be  gen- 
erated by  revolving  the  figure  ACDB  about 
AO  as  an  axis. 

We  have  Vol.  generated  by  ODB  =  —  r*h.  §  375 

3 

Vol.  generated  by  OAB  =  -r^(h  +  d).         (Why  ?) 

3 


Vol.  generated  by  OOD  =  -r<?d.  (Why  ?) 

o  •      , 

Hence,  v  =  ^r2A  +  |n»(ft  +  d)-  ^rjd 

=  1  [2  r2/*  +  n%  +  d(rj  -  r,*)].  (1) 

o 

From    r2  =  r22  +  d2  and  r2  =  rx2  +(h  +  d)* 

we  obtain  d  =  TZ  ~TI  ~~ — .  (2) 

2  h 

Substituting  this  value  of  d  in  r2  =  r22  +  d2/ 


we  get      r.  =  «•«'  +  n4  +  V  -  2  W  +  2  *T*  +  2  ^  (3) 

Substituting  (2)  and  (3)  in  (1)  and  reducing,  we  have 


*-•'• 


377,   COROLLARY.     The  volume  of  a  spherical  segment  of  one 
base  is 


Suggestion.     In  this  case  n2  =  0  and  r22  =  r2  —  (r  —  &)2  =  2  r^  -  #>. 
Substituting  in  the  formula  above,  we  have  the  result. 


THE  SPHERE  147 


SUMMARY  OF  BOOK  V 

1.  Define  sphere,  diameter,  radius. 

2.  Collect  the  theorems  of   Book  V  involving   plane   sec- 
tions of  the  sphere. 

3.  Define  axis  and  pole  of  a  circle,  great  circle,  polar  distance. 

4.  Define  tangent  plane  to  a  sphere,  inscribed  and  circum- 
scribed polyhedrons. 

5.  Arrange  in  parallel  columns  the  corresponding  theorems 
on  trihedral  angles  and  spherical  triangles   which  are  proved 
without  the  use  of  polar  triangles. 

6.  Collect  the  definitions  on  polar  triangles. 

7.  Collect  the  theorems  on  polar  triangles. 

8.  Continue   the   lists  begun  in  Example  5,   adding  the 
theorems  proved  by  means  of  polar  triangles. 

9.  Make  a  list  of  the  definitions  involving  polyhedral  angles 
and  spherical  polygons. 

10.  Collect   the   theorems  involving  polyhedral  angles  and 
spherical  polygons. 

11.  Collect  the  theorems  on  the  areas  of  spherical  triangles 
and  polygons. 

12.  Give  the  definitions  and  assumptions  pertaining  to  the 
area  and  volume  of  the  sphere. 

13.  State  all  the  theorems  pertaining  to  the  area  and  volume 
of  the  sphere. 

14.  Give  the  definitions  and  theorems  pertaining  to  spherical 
figures,  such  as  zones,  cones,  sectors,  segments. 

15.  Collect  all  the  mensuration  formulas  in  this  Book. 

16.  Collect  all  the  mensuration  formulas  of  solid  geometry. 

17.  Describe  some  of  the  most  important  applications   in 
this    Book.      Return    to    this   question   after   studying    the 
following  sets  of  problems. 


148  SOLID   GEOMETRY:    BOOK  V 

PROBLEMS  ON  BOOK  V 

1.  What  part  of  the  earth's  surface  lies  in  the  torrid  zone  ? 
What  part  in  the  temperate  zones  ?     What  part  in  the  frigid 
zones  ?     The  parallels  23^°  north  and  south  of  the  equator 
are  the  boundaries  of  the  torrid  zone,  and  the  parallels  66|° 
north  and  south  are  the  boundaries  of  the  frigid  zones. 

2.  Find  to  four  places  of   decimals  the  area  of  a   sphere 
circumscribed  about  a  cube  whose  edge  is  6.     No  square  root 
is  to  be  approximated  in  the  process,  and  the  value  of  TT   is 
taken  as  3.1416. 

3.  Can  the  volume  of  the  sphere  in  the  preceding  exercise 
be  approximated  without  finding  a  square  root?     Find  the 
volume. 

4.  Find  the  area  of  a  sphere  circumscribed  about  a  rectangu- 
lar parallelepiped  whose  sides  are  a,  6,  and  c. 

5.  Find  the  volume  of  the  sphere  in  the  preceding  example. 

6.  A  fixed  sphere  with  center  0  has  its  center  on  another 
sphere  with  center  0'.     Show  that  the  area  of  the  part  of  O' 
which  lies  within  0  is  equal  to  the  area 

of  a  great  circle  of  the  sphere  0,  pro- 
vided the  radius  of  the  sphere  0  is  not 
greater  than  the  diameter  of  0'. 

Suggestion.     Let  the  figure  represent  a  cross 
section  through  the  centers  of  the  two  spheres. 

Connect  0  with  A  and  B.  Then  OA2  =  OB  x  OD.  But  OD  is  the 
altitude  of  the  zone  of  O1  which  lies  within  0,  and  OB  is  the  diameter 
of  the  sphere  O'.  Hence,  the  area  of  the  zone  is  IT  OB  x  OD  =  irOA2. 

7.  Given  a  solid  sphere  of  radius  12  inches.     A  cylindrical 
hole  is  bored  through  it  so  that  the  axis  of  the  cylinder  passes 
through  the  center  of  the  sphere.     What  area  of  the  sphere 
is  removed  if  the  diameter  of  the  hole  is  4  inches  ? 

8.  Find  the  volume  removed  from  the  sphere  by  the  process 
described  in  the  preceding  exercise. 


THE  SPHERE  149 


9.  A  cylindrical  post  6  in.  in  diameter  is  surmounted  by  a 
part  of  a  sphere  10  in.  in  diameter,  as  shown  in  the  figure. 
Find  the  surface  and  the  volume  of  the  part  of 
the  sphere  used. 

10.  A  cylindrical  post  5  ft.  long  and  4  in.  in 
diameter  is  surmounted  by  a  part  of  a  sphere  9 
in.  in  diameter  as  shown  in  the  figure.     Find 
the  volume  of  the  whole  post   including   the 
spherical  part. 

11.  Find  the  volume   of  a  spherical    shell 
one  inch  thick  if  its  outer  diameter  is  8  inches. 

12.  Compare  the  volumes  and  areas  of  a  sphere  and  the  cir- 
cumscribed cylinder. 

13.  In  a  sphere  of  radius  r  a  cylinder  is  inscribed  whose  alti- 
tude is  equal  to  its  diameter.     Compare  its  volume  and  area 
with  those  of  the  sphere. 

14.  Find  the  diagonal  of  a  cube  in  terms  of  its  side,  and  also 
a  side  in  terms  of  half  the  diagonal. 

15.  Express  the  volume  of  a  cube  inscribed  in  a  sphere  in 
terms  of  the  radius  of  the  sphere. 

16.  Three  spheres  each  of  radius  r  are  placed  on  a  plane  so 
that  each  is.  tangent  to  the  other  two.     A  fourth  sphere  of 
radius  r  is  placed  on  top  of  them.     Find  the  distance  from  the 
plane  to  the  top  of  the  upper  sphere. 

17.  Find  the  vertical  distance  from 
the  floor  to  the  top  of  a  triangular 
pile  of  spherical  cannon  balls,  each 
of   radius    5  inches,   if  there  are   3 
layers  in  the  pile. 

18.  Solve  a  problem  like  the  pre- 
ceding if  there  are  16  layers  in  the 
pile,  each  shot  of  radius  r. 


150  SOLID  GEOMETRY:    BOOK  V 

FORMULAS  DEVELOPED  IN  BOOKS  IH,  IV,  V 

1.  If    V  is   the   volume   of    a  rectangular    parallelepiped 
whose  dimensions  are  a,  6,  c,  then 

V=abc. 

2.  If  V  is  the  volume,   b  the  area  of  the  base,  and  h  the 
altitude  of  a  prism  or  cylinder,  then 

V=hb. 

3.  If  S  is  the  lateral  surface,  p  the  perimeter  of  a  right 
section  of  a  prism  or  cylinder,  and  e  the  lateral  edge  or  ele- 
ment, then 

S=pe. 

4.  If  V  is  the  volume,  b  the  area  of  base,  and  h  the  altitude 
of  a  pyramid  or  cone,  then 


5.   If  $  is  the  lateral  area,  p  the  perimeter  of  the  base,  and 
I  the  slant  height  of  a  regular  pyramid  or  cone,  then 


6.    If  V  is  the  volume,  b  the  lower  base,  b'  the  upper  base, 
and  h  the  altitude  of  a  frustum  of  pyramid  or  cone,  then 


7.   If  S  is  the  area  of  the  surface,  V  the  volume,  and  r  the 
radius  of  a  sphere,  then 


8.  If  a  is  the  spherical  excess  in  degrees  of  a  spherical 
polygon,  S  the  area  in  square  units,  and  r  the  radius  of  a 
sphere,  then 


9.   If  S  is  the  area,  h  the  altitude,  and  r  the  radius  of  a  lune, 
then 


THE  SPHERE  151 


10.   If  V  is  the  volume  of   spherical  cone,  h  the  altitude, 
and  r  the  radius  of  the  sphere,  then 


11.  If  V  is  the  volume  of  a  spherical  segment,  r  the  radius 
of  the  sphere,  rt  and  r2  the  radii  of  the  bases,  and  h  the  altitude 
of  the  segment,  then 


SUPPLEMENTARY  EXERCISES  IN  COMPUTATION 

1.  Find  the  volume  of  a  rectangular  parallelepiped  if 
(1)  a  =  6  (2)  a  =  f  (3)  a  =  V3  (4)  a  =  3\/5 

6=8  &  =  T9<r  6  =  V6  6  =  Vl5 

c  =  7f  c  =  3  c  =  4  c  =  3 

2.  Find  volume  of  a  prism,  or  cylinder,  if 

(1)6  =  12  (2)  6=V3  (3)6  =  4V2  (4)6  = 

h  =  16  h=V6  h 


3.  Find  volume  of  a  pyramid  or  cone  if 

1)  6  =  46  (2)  6  =  47£  (3)  6  =  34  I 

A  =  12  /i  =  10  A  =  8 

4.  Find  the  lateral  surface  of  a  regular  pyramid  or  cone  if 


(1)  6  =  46  (2)  6  =  47£  (3)  6  =  34  IT  (4)  6  =  42  ir 

A  =  12  /i  =  10  A  =  8  A  =  6  V2 


5.  Find  the  volume  of  a  frustum  of  a  pyramid  or  cone  if 
(1)6  =  8*  (2)6=2V2          (3)6  =  167r  (4)   6  =  3  V2 

6'  =  6  6'=V3  6'=97r  6' 

h  =  4  h  =  6  A  =  3  h 

6.  Find  the  volume  or  surface  of  a  sphere  if 

r  =  4,  r  =  8,  r  =  3^2,  r  =  7  V3,  r  =  12V6~ 


7.   Find  the  area  of  a  spherical  polygon  if 
(1)  a  =  84°  (2)  a  =112°  30'  (3)  a  =  49°  25'  17" 

r  =  12  r  =  8  r  =  14 


152  SOLID  GEOMETRY:    BOOK   V 

MISCELLANEOUS   REVIEW  EXERCISES 
A.   Locus  PROBLEMS 

1.  Find  the  locus  of  all  points  in  space  equally  distant  from 
two  parallel  lines  and  also  from  two  parallel  planes.     Discuss. 

2.  Find  the  locus  of  all  points  in  space  equally  distant 
from  each  of  two  intersecting  straight  lines  and  also  from  two 
intersecting  planes.     Discuss. 

3.  What  is  the  locus  of  all  points  at  a  perpendicular  dis- 
tance of  2  feet  from  a  given  line  and  lying  in  a  plane  parallel 
to  the  line  ?     Discuss. 

4.  Find  the  locus  of  all  points  equidistant  from  two  given 
points  A  and  B,  and  also  equidistant  from  two  planes  M  and 
N.     Discuss. 

5.  What  is  the  locus  of  all  points  on  the  floor  of  a  room 
which  are  equally  distant  from  two  diagonally  opposite  corners 
of  the  room,  one  in  the  floor  and  one  in  the  ceiling  ? 

6.  Find  the  locus  of  all  points  on  a  sphere  where  it  is  met 
by  line-segments  of  equal  length  drawn  from  a  fixed  point  P 
outside  the  sphere.     Discuss. 

7.  Find  the  locus  of  all  points  which  are  at  the  same  fixed 
distance  from  each  of  two  intersecting  planes,  M  and  JV,  and 
also  equally  distant  from  two  planes,  P  and  Q. 

8.  Given  a  plane  M  and  a  point  P  not  in  M.     Find  the 
locus  of  a  point  which  divides  in  a  given  ratio  each  segment 
connecting  P  with  a  point  in  M :    (a)  if  the   segments   are 
divided  internally;  (6)  if  they  are  divided  externally. 

9.  A  segment  AB  of  fixed  length  is  free  to  move  so  that 
its  end-points  lie  in  two  fixed  parallel  planes.     Find  the  locus 
of  a  point  C  on  AB  if  AC  is  of  fixed  length. 

10.  Given  two  fixed  points  in  space,  through  each  of  which 
passes  a  system  of  straight  lines.  If  each  line  of  one  system 
is  perpendicular  to  a  line  of  the  other  system,  find  the  locus 
of  the  intersection  points. 


THE  SPHERE 


153 


B.   PROBLEMS  IN  NUMERICAL  COMPUTATION 

1.  A  pedestal  for  a  monument  is  in  the  shape  of  a  frustum 
of  pa  regular  hexagonal  pyramid,  the  radius  of  the  upper  base 
being  4  ft.,  that  of  the  lower  base  6  ft.,  and  the  altitude  of  the 
frustum  8  ft.    Find  its  volume,  slant  height,  and  lateral  surface. 

2.  The  area  of  the  lower  base  of  a  frustum  of  a  pyramid  is 
42  sq.  ft.,  its  altitude  8  ft.,  and  volume  200  cu.  ft.     Find  the 
area  of  the  upper  base. 

3.  The  area  of  the  base  of  a  pyramid  is  480  sq.  ft.  and  its 
altitude  30  ft.     Find  the  volume  of  the  frustum  remaining 
after  a  pyramid  with  altitude  10  ft.  has  been  cut  off  by  a  plane 
parallel  to  the  base. 

4.  The  area  of  the  base  of  a  pyramid  is  250  sq.  in.     If  a 
plane  section  of  the  pyramid  parallel  to  the  base  and  at  a  dis- 
tance of  5  in.  from  it  has  an  area  of  175  sq.  in.,  find  the  alti- 
tude of  the  pyramid. 

5.  The  figure  below  represents  a  solid  whose  base  is  a  rec- 
tangle 50ft.  long  and  40  ft.  wide.     Its  height  is  12  ft.  and  its 
top  a  rectangle  20  ft.  by  10  ft.     Find  its  volume. 


15 


15 


B 


20 


15 


6.  A  frustum  of  a  right  circular  cone  has  an  altitude  one 
half  that  of  the  cone.     If  its  slant  height  is  8  ft.  and  lateral 
area  64  IT  sq.  ft.,  find  the  diameters  of  its  bases. 

7.  If  the  area  of  the  base  of  a  cone  is  16?r  sq.  in.  and  its 
altitude  6  in.,  find  the  distance  from  the  vertex  to  a  plane, 
parallel  to  the  base,  which  cuts  out  a  section  of  area  9?r  sq.  in. 


154  SOLID  GEOMETRY:    BOOK  V 

C.   PROBLEMS  IN  ALGEBRAIC  COMPUTATION 

1.  Find  the  total  area  and  the  volume  of  a  regular  tetra- 
hedron each  of  whose  edges  is  e. 

2.  If  the  numerical  values  of  the  volume  and  of  the  total 
area  of  a  regular  tetrahedron  are  equal,  what  is  the  length  of 
its  edge  ? 

3.  Find  the  length  of  an  edge  of  a  regular  tetrahedron  if 
its  volume  is  numerically  equal  to  the  square  of  the  edge. 

4.  Cut  a  pyramid  of  altitude  h  by  means  of  a  plane  parallel 
to  the  base  so  that  the  perimeter  of  the  section  shall  be  one 
third  that  of  the  base. 

5.  If  the  altitude  of  a  pyramid  is  h,  how  far  from  the  base 
must  a  plane  parallel  to  it  be  drawn  so  that  the  area  of  its 
cross  section  shall  be  half  that  of  the  base  of  the  pyramid  ? 

6.  In  a  regular  right  pyramid  a  plane  parallel  to  the  base 
cuts  it  so  as  to  make  a  section  whose  area  is  one  half  that  of 
the  base.     Find  the  ratio  between  the  lateral  area  of  the  pyra- 
mid and  that  of  the  small  pyramid  cut  off  by  the  plane. 

7.  If  the  diameter  of  a  right  circular  cylinder  is  equal  to 
its  altitude,  determine  the  diameter  so  that  the  total  area  of 
the  cylinder  shall  be  equal  numerically  to  its  volume. 

8.  Cut  a  right  circular  cone  of  altitude  h  by  a  plane  parallel 
to  the  base  so  that  the  area  of  the  section  shall  be  one  third 
that  of  the  base.     Find  the  distance  from  the  vertex  to  the  plane. 

9.  Show  that  the  lateral  area  of  the  small  cone  cut  off  in 
Example  8  is  one  third  the  lateral  area  of  the  original  cone. 

10.  In  a  right  circular  cone,  with  alti- 
tude h,  and  r-  the  radius  of  its  base,  a  cylin- 
der is  inscribed  as  shown  in  the  figure. 
Find  the  radius  OF  of  the  cylinder  if  the 
area  of  the  ring  bounded  by  the  circles  OF 
and  OA  is  equal  to  the  lateral  area  of  the 
small  cone  cut  off  at  the  top. 


THE  SPHERE  155 


D.   PROBLEMS  IN  CONSTRUCTION 

1.  Construct  a  plane  tangent  to  a  given  sphere  and  parallel 
to  a  given  plane.     How  many  such  planes  are  there  ? 

Suggestion.  From  the  center  of  the  sphere  pass  a  line  perpendicular 
to  the  given  plane  and  draw  tangent  planes  at  the  points  where  this  line 
meets  the  surface  of  the  sphere.  See  §§  306,  129. 

2.  Construct  a  plane  tangent  to  a  given  sphere  and  per- 
pendicular to  a  given  line.     How  many  such  planes  are  there  ? 

3.  How  many  planes  may  be  tangent  to  a  sphere  at  a  point 
on  the  sphere?     How  many  lines?     Show  how  to  construct 
them. 

4.  Through  a  given  point  exterior  to  a  sphere  construct  a 
line  tangent  to  the  sphere. 

5.  How  many  lines  tangent  to  a  sphere  can  be  constructed 
from  a  point  outside  the  sphere  ? 

6.  Through  a  given  point  exterior  to  a  sphere  construct  a 
plane  tangent  to  the  sphere.     How  many  planes  can  be  passed 
through  a  given  exterior  point  tangent  to  the  sphere  ? 

7.  How  many  planes  tangent   to  a  sphere  can  be  passed 
through  two  given  points  A  and  B  outside  a  sphere  ?     Dis- 
cuss fully  if  the  line  AB  (1)  meets  the  sphere  in  two  points ; 
(2)  is  tangent  to  the  sphere ;    (3)  does  not  meet  the  sphere. 
Show  how  to  make  the  constructions. 

8.  Is  it  possible  to  construct  a  spherical  triangle  each  of 
whose  angles  is  a  right  angle  ?     Show  how  to  construct  one. 
In  a  trirectangular  spherical  triangle  what  is  the  length  of 
each  side  in  terms  of  degrees  ? 

9.  Construct  a  spherical  triangle  such  that  its  polar  triangle 
is  identical  with  the  given  triangle. 

10.  Construct  a  spherical  triangle  whose  sides  are  70°,  80°, 
and  110°,  respectively,  and  find  the  sides  of  each  of  the  eight 
triangles  formed  by  its  polar  construction. 


156  SOLID  GEOMETRY:    BOOK  V 

E.   THEOREMS  TO  BE  PROVED 

1.  Prove  that  planes  perpendicular  to  the  faces  of  a  tri- 
hedral angle  and  bisecting  its  face  angles  meet  in  a  line. 

2.  Prove  that  if  in  two  tetrahedrons  three  faces  of  one  are 
equal  respectively  to  three  faces  of  the  other  and  similarly 
placed  about  a  vertex,  the  tetrahedrons  are  equal. 

3.  Prove  that  two  tetrahedrons  are  equal  if  two  faces  and 
the  included  dihedral  angle  are  equal  and  similarly  placed. 

4.  In  any  regular  tetrahedron,  an  altitude  equals  three  times 
the  perpendicular  from  its  foot  to  any  face ;   or  an  altitude 
equals  the  sum  of  the  perpendiculars  to  the  faces  from  any 
point  within  the  tetrahedron. 

5.  In  a  cylinder  of  revolution  the  diameter  of  whose  base 
equals  the  altitude,  the  volume  equals  one  third  the  product  of 
the  total  surface  by  the  radius  of  the  base. 

6.  If  two  intersecting  planes  are  each  tangent  to  a  cylinder, 
show  that  their  line  of  intersection  is  parallel  to  an  element 
of  the  cylinder  and  also  parallel  to  the  plane  containing  the 
two  elements  of  contact. 

7.  The  volume  of  a  frustum  of  a  right  circular  cone  equals 
the  sum  of  a  cylinder  and  a  cone  of  the  same  altitude  as  the 
frustum,  and  with  radii  which  are  respectively  the  half  sum 
and  the  half  difference  of  the  radii  of  the  frustum. 

8.  Of  circles  on  a  sphere  whose  planes  pass  through  a  given 
point  within  the  sphere,  the  smallest  is  that  circle  whose  plane 
is  perpendicular  to  the  diameter  through  the  given  point. 

9.  A  right  triangular  prism  is  cut   by  a 
plane  not   parallel   to   the  base,   but   such 
that  its  intersection  DE  is  parallel  to  the 
base   segment  AB.     Show  that  the  volume 
of  the  part  thus  cut   off   is   one   third  the 
product  of  the   sum   of  the   three   vertical 
edges  and  the  area  of  the  base. 


Adrien  Marie  Legendre  (1752-1833)  was  a  celebrated  French  mathe- 
matician. He  was  one  of  three  commissioners  who  introduced  the 
metric  system  in  France,  having  also  been  a  member  of  the  commis- 
sion for  determining  the  length  of  the  meter. 

Besides  many  treatises  on  advanced  subjects,  he  wrote  a  book  on 
elementary  geometry  which  was  the  most  successful  of  the  many 
attempts  to  supersede  Euclid  as  a  text-book.  It  went  through  many 
editions  in  French  and  was  translated  into  almost  every  other  civilized 
language. 


APPENDIX 


APPENDIX  I  :    SIMILAR  SOLIDS 

378,  Similar   Cylinders.     Two   right    circular   cylinders   are 
similar  if  they  are  generated  by  similar  rectangles  revolving 
about  corresponding  sides. 

379,  THEOREM  I.     If  in  two  similar  cylinders  s  and 
s  are  the  lateral  areas,  S  and  S'  the  total  areas,  V  and 
F'  the  volumes,  h  and  h'  the  altitudes,  and  r  and  r'  the 
radii,  then 

1  =  ^==^-  =  ^     d  Y  -^=^. 

' 


s' 


h'* 


\h 


..r-f. 


Suggestions  for  Proof :    To  find  the  ratios   —  and  — ,  make 

s  o 

use  of  the  following,  giving  reasons  for  each  in  detail. 

h    __j  /^\    r  +  h    _^__7i 
=  r'~hr 


To  find  the  ratio  —    make  use  of  F=  irr2h,  and  V  =  irr'2h', 


H 


157 


158 


SOLID   GEOMETRY:    APPENDIX  I 


RATIOS  RELATING  TO  SIMILAR  CONES 

380,  Similar  Cones.     Two  right  circular  cones  are  similar  if 
they  are  generated  by  two  similar  right  triangles  revolving 
about  corresponding  sides. 

381,  THEOREM  II.     If  in  two  similar  cones  s  and  s' 
are  the  lateral  areas,  S  and  S'  the  total  areas,  V  and 
V  the  volumes,  h  and  h'  the  altitudes,  and  r  and  r' 
the  radii,  then 


-  =  -  =  —  =     -  and  -=      =      . 


Suggestions  for  Proof :    See  suggestions  under  §  379. 

EXERCISES 

1.  The  lateral  area  of  a  cone  is  36  square  inches.     What  is 
the  lateral  area  of  a  similar  cone  whose  altitude  is  f  that  of 
the  given  cone  ? 

2.  The  total  area  of  one  of  two  similar  cones  is  three  times 
that  of  the  other.     Compare  their  altitudes  and  also   their 
radii. 

3.  The  sum  of  the  total  areas  of  two  similar  cones  is  144 
square  inches.     Find  the  area  of  each  cone  if  one  is  If  times 
as  high  as  the  other. 

4.  The  volume  of  one  of  two  similar  cones  is  5  times  that 
of  the  other.     Compare  their  altitudes. 


SIMILAR   SOLIDS 


159 


CONDITIONS  FOR  SIMILARITY  OP  TETRAHEDRONS 

Similar  Polyhedrons.  Two  polyhedrons  are  similar  if 
they  have  the  same  number  of  faces  similar  each  to  each  and 
similarly  placed,  and  have  their  corresponding  polyhedral  angles 
equal. 

Any  two  parts  which  are  similarly  placed  are  called  corre- 
sponding parts,  as  corresponding  faces,  edges,  vertices. 

383,  THEOREM  III.  Two  tetrahedrons  are  similar 
if  three  faces  of  one  are  similar  respectively  to  three 
faces  of  the  other,  and  are  similarly  placed. 


Given  the  tetrahedrons  P-ABC  and  Pf-A'BfCf  having 
A  APB  ~  A  A'P'K,  A  APC  ~  A  A'PC,  and  A  BPC  - 

To  prove  P-ABC  ~  P'-A'B'C'. 

Proof :  (1)  Show  that  A  ABC  ~  A  A'B'C'. 

(2)  Show  that  trihedral  A  P  and  P'  are  equal. 

Likewise  Z  J.  =  Z  ^',  Z  5  =  Z  5',  Z  (7  =  Z  (7. 

Hence,  by  definition,  the 
polyhedrons  are  similar. 

SIGHT  WORK 

1.  If  the  two  prisms  in  the  figure 
are  similar,  name  the  pairs  of  corre- 
sponding parts.  Likewise  for  two 
similar  pyramids. 


AB'P'C. 


160 


SOLID  GEOMETRY:    APPENDIX  I 


TETRAHEDRONS  HAVING  TWO  EQUAL  TRIHEDRAL  ANGLES 

384,  THEOREM  IV.  The  volumes  of  two  tetrahe- 
drons, having  a  trihedral  angle  of  the  one  equal  to  a 
trihedral  angle  of  the  other,  are  proportional  to  the 
products  of  the  edges  which  meet  in  the  vertices  of 
these  angles. 


Given  the  tetrahedrons  P-ABC  and  Pf-AfBfCr  whose  volumes 
are  V  and  V  and  in  which  Tri.  Z  P  =  Tri.  Z  P1. 


Proof:  Place  P'-A'B'C'  so  that  Tri.  ZP'  coincides  with 
Tri.  Z  P. 

Let  CM  and  C'M'  be  the  altitudes  of  P-ABC  and  P-A'B'C' 
from  the  vertices  C  and  C'  upon  the  plane  PAB. 

Let  A2Vand  A'N'  be  the  altitudes  of  the  A  PAB  and  PA'B'. 


Th       V 


CM  •  PB  •  AN 


V     \  C'M'  •  area  PA'B'      C'M'  -  PB'  -  A'N' 


——  >-    - 


Give  all  the  steps  and  reasons  in  detail. 


SIMILAR   SOLIDS  161 


VOLUMES  OF  SIMILAR  TETRAHEDRONS 

385,  THEOREM  V.  The  volumes  of  two  similar 
tetrahedrons  are  in  the  same  ratio  as  the  cubes  of  their 
corresponding  edges. 

Given  P-ABC  ~  P'-A'B'C',  with  volumes  V  and  Y. 

V      PA3 
To  prove  that  —  =  . 

r  P*'  A'6 

Proof :  We  have  —  =  ^  Al  '         '  ^,^t .  §  384 


Now  use  the  properties  of  similar  triangles  to  complete  the 
proof.     Use  the  figure  of  §  384. 

EXERCISES 

1.  Show  that  two  tetrahedrons  are  similar  if  they  have  a 
dihedral  angle  of  one  equal  to  a  dihedral  angle  of  the  other 
and  the  including  faces  similar  each  to  each  and  similarly 
placed. 

2.  Show  that  the  total  areas  of  two  similar  tetrahedrons  are 
in  the  same  ratio  as  the  squares  of  any  two  corresponding 
edges. 

3.  Show  that  if  each  of  two  polyhedrons  is  similar  to  a 
third  they  are  similar  to  each  other. 

4.  Two  similar  tetrahedral   mounds  have  a  pair  of  corre- 
sponding dimensions  3  ft.  and  4  ft.     If  one  mound  contains 
40  cu.  ft.  of  earth,  how  much  does  the  other  contain  ? 

5.  The  edges  of  a  tetrahedron  are  3,  4,  5,  6,  7,  and  8.     Find 
the  edges  of  a  similar  tetrahedron  containing  64   times   the 
volume. 

6.  Find  what  fraction  of  the  altitude  of  a  tetrahedron  must 
be  cut  off  by  a  plane  parallel  to  the  base,  measuring  from  the 
vertex,  in  order  that  the  new  pyramid  thus  detached  may  have 
one  fifth  of  the  original  volume. 


162 


SOLID   GEOMETRY:    APPENDIX  I 


FIGURES  HAVING  A  CENTER  OF  SIMILITUDE 

386,  Center  of  Similitude.     Two  figures  are  said  to  have  a 
center  of  similitude  O,  if  for  any  two  points,  A  and  B,  of  the 
one  figure  the  lines  AO  and  BO  meet  the  other  in  two  points, 
A'  and  B',  called  corresponding  points,  such  that 

OA  :  OA  =  OB  :  OB1. 
See  figures  under  §§  387-392. 

387.  THEOREM  VI.     Any  two  figures  which  have  a 
center  of  similitude  are  similar. 

Proof  :  (1)   Two  triangles. 


Given 


OA'     OB'     OO 


Let  the  student  prove  that  A  ABC  ~  A  A'B'C'. 
In  case  the  triangles  do  not  lie  in  the  same  plane,  use  §  101 
to  show  that  the  corresponding  A  are  equal. 
(2)   Two  polygons. 

Given        ^=H'=l§'ete- 


Give  the  proof  both  for  polygons  in  the  same  plane  and 
not  in  the  same  plane. 


SIMILAR   SOLIDS  163 


(3)   Two  tetrahedrons. 


With  the  same  hypothesis  as  before,  we  must  prove 
A  PAB  ~  P'A'B',  A  PBC  ~  A  P'B'C',  A  PGA  ~  A  POA',  and 
then  use  §  383. 

(4)  Any  two  polyhedrons. 

(a)  Prove  corresponding  polygonal  faces  similar. 

(6)  Prove  corresponding  polyhedral  angles  equal. 

The  last  step  requires  not  only  equal  face  angles  about  the 
vertex,  as  in  the  case  of  the  tetrahedron,  but  also  equal  dihedral 
angles.  Note  that  two  dihedral  angles  are  equal  if  their  faces 
are  parallel  right  face  to  right  face  and  left  face  to  left  face. 
(Why?) 

(5)  Consider  any  two  figures  whatsoever  having  a  center  of 
similitude. 

(a)  Take  any  three  points  A,  B,  C,  in  one  figure  and  the 
three  corresponding  points,  A',  B',  <7',  in  the  other. 

Then  AB  and  A'B',  AC  and  A'C',  etc.,  are  called  corre- 
sponding linear  dimensions,  and  the  triangles  ABC  and  A'B'C' 
are  corresponding  triangles. 

(6)  It  is  clear  that  any  two  corresponding  linear  dimensions 
have  the  same  ratio  as  any  other  two,  and  that  any  two  cor- 
responding triangles  are  similar. 

In  this  sense  any  two  figures  having  a  center  of  similitude 
are  thus  proved  to  be  similar. 


164  SOLID   GEOMETRY:    APPENDIX  I 

SIMILAR  TRIANGLES   PLACED  IN   SIMILITUDE 

388,  Ratio   of   Similitude.     The   ratio   of  similitude   of  two 
similar   figures  is  tlie  common   ratio"  of   their   corresponding 
linear  dimensions.     This  ratio  is  the  same  as  the  distance  ratio 
of  corresponding  points  from  the  center  of  similitude. 

389.  THEOREM  VII.     Two  similar  triangles  may  be 
so  placed  as  to  have  a  center  of  similitude. 


Given  the  similar  triangles  T  and  7y,  in  which 
A'B     A'C 


AB       AC       EC  ' 

To  prove  that  they  may  be  placed  with  a  center  of  similitude. 
Proof  :  From  any  point  0  draw  OA,  OB,  OC. 
On  these  rays  take  Al}  B^  Ci  so  that 

OAl=OB1=OCl  =  A'B' 
OA       OB       OC      AB' 

Now  show  the  following  : 

(1)  A  2\  ~  A  T,  and  hence  A  2^  ~  A  T'. 

(2)  A  T!  =  A  F. 

For  this  show  that  A^BV  —  A'B'  by  means  of  the  equations 


d 

' 


A'B'_OA1 


AB  ™  OA  AB  ~~  OA 


SIMILAR   SOLIDS 


165 


Likewise  Ai 
(3)  Finally, 


and  1 

ON, 
OM 


B'C'. 


OAl 
~OA' 


where  M  and  MI  are  any  two  corresponding  points. 
Hence  0  is  the  required  center  of  similitude. 

SIMILAR  TETRAHEDRONS  PLACED  IN  SIMILITUDE 

390,    THEOKEM  VIII.     Two  similar  tetrahedrons  may 
be  so  placed  as  to  have  a  center  of  similitude. 


Given  the  similar  tetrahedrons  T  and  T'. 

To  prove  that  they  can  be  placed  so  as  to  have  a  center  of 
similitude. 

Proof:  With  0  as  a  center  of  similitude,  construct  2\, 
making 


OAl     OB, 


A'B' 


Now  show  as  in  §  389  that  2\  =  Tf,  and  hence  that  T  can 
be  placed  in  the  position  2\  so  as  to  have  with  T  the  center 
of  similitude  0. 

Give  all  the  steps  in  detail. 

391.  COROLLARY.  Any  two  similar  polyhedrons  may  be 
placed  so  as  to  have  a  center  of  similitude. 

Suggestion.  The  argument  is  precisely  similar  to  that  in  §  390. 


166 


SOLID   GEOMETRY:    APPENDIX  I 


CENTER   OF  SIMILITUDE   WITHIN   THE   FIGURE 

392.  In  the  proofs  for  the  two  preceding  theorems  the  center 
of  similitude  was  taken  between  the  two  figures  or  on  the  same 
side  of  them.  The  center  may  be  taken  equally  well  within 
them,  as  in  the  following  illustrations : 


B' 


FIG.  1 


In  the  case  of  similar  convex  polyhedrons  with  the  center 
of  similitude  thus  placed,  the  faces  are  the  bases  of  pyramids 
whose  vertices  are  all  at  the  center  of 
similitude. 

If,  further,  the  polygonal  faces  be 
divided  into  triangles  by  drawing  their 
diagonals,  these  triangles  become  the 
bases  of  tetrahedrons,  all  of  whose  ver- 
tices are  at  the  center  of  similitude. 

Moreover,  each  inner  tetrahedron  is 
similar  to  its  corresponding  outer  tetra- 
hedron. (Why?) 

The  volumes  of  the  two  similar  polyhedrons  are  thus  com- 
posed of  the  sums  of  sets  of  similar  tetrahedrons. 

SIGHT  WORK 

1.  Give  the  proof  of  §  389,  using  Fig.  1  above,  and  extend  the  argu- 
ment to  two  similar  polygons,  using  Fig.  2. 

2.  Draw  two  similar  tetrahedrons  with  their  center  of   similitude 
within  them,  and  give  the  proof  of  §  390  with  the  figure  thus  made. 

3.  Give  the  proof  of  §  390,  using  Fig.  3  above. 


FIG.  3 


SIMILAR   SOLIDS  167 


RATIO  OF  VOLUMES  OF  SIMILAR  POLYHEDRONS 

393,  THEOREM  IX.  The  volumes  of  any  two  similar 
polyhedrons  are  proportional  to  the  cubes  of  their  cor- 
responding edges. 

Proof:  Place  the  polyhedrons  whose  volumes  are  Fand  V 
so  as  to  have  their  center  of  similitude  within  them  as  in  the 
third  figure  of  §  392. 

Call  the  volumes  of  the  similar  tetrahedrons  2\,  T2,  T3,  •••, 
and  2*!,  2*2,  2*3,  —  ,  and  let  AB  and  A'B'  be  two  correspond- 
ing edges. 

Then  we  have 


n   *%   2*3 

And  it.      -'  =lL  =         ~  (Why  ? 


But  2\  +  T2 +  273  .»  =  Fand  Tl  +  T'2  +  2*3  »•  =  F'. 

Hence,  t, 


394,  COROLLARY.  The  volumes  of  any  two  similar  solids  are 
proportional  to  the  cubes  of  any  two  corresponding  linear 
dimensions. 

This  proposition  may  be  rendered  evident  by  noticing  that 
any  two  similar  three-dimensional  figures  may  be  built  up  to 
any  degree  of  approximation  by  means  of  pairs  of  similar 
tetrahedrons  similarly  placed.  The  proposition  then  holds 
for  any  two  corresponding  figures  used  in  the  process  of 
approximation. 

Note  that  the  ratio  of  similitude  of  two  similar  figures  may 
be  obtained  from  the  ratio  of  any  pair  of  their  corresponding 
linear  dimensions. 


168 


SOLID   GEOMETRY:    APPENDIX  I 


APPLICATIONS  OP  SIMILARITY 

395,  The  Pantograph.  The  theorem  that  any  two  figures 
which  have  a  center  of  similitude  are  similar  is  the  geometric 
basis  of  many  mechanical  contrivances  for  enlarging  or  reduc- 
ing both  plane  and  solid  figures ;  that  is,  for  constructing  fig- 
ures similar  to  given  figures 
and  having  with  them  a  given 
ratio  of  similitude. 

The  essential  property  of  all 
such  contrivances  is  that  one 
point  O  is  kept  fixed,  while 
two  points  A  and  B  are  allowed 
to  move  so  that  0,  A,  and  B 
always  remain  in  a  straight 
line,  and  so  that  the  ratio 
OA  :  OB  remains  the  same.  See  page  216  of  Plane  Geometry. 

In  the  first  figure  on  this  page  0  is  a  fixed  point.  Seg- 
ments OD,  CB,  and  the  sides  of  the  parallelogram  AGED  are 
of  fixed  length. 

Prove  that  if  B  is  once  so  taken  on  the  line  EC  as  to  be 
in  the  line  OA,  the  points  0,  A,  and  B  will  always  remain 
collinear,  and  that  OA :  OB  re- 
mains a  fixed  ratio. 

In  the  second  figure  is  shown 
an  ordinary  pantograph  used 
for  copying  and  at  the  same 
time  for  reducing  or  enlarging 
maps,  designs,  etc.  The  lengths 
of  the  various  segments  are 
adjustable,  as  shown,  thus  ob- 
taining any  desired  scale. 

The  same  contrivance  may  be  used  for  copying  figures  in 
space,  such  as  relief  maps,  and  at  the  same  time  reducing  or 
enlarging  them. 


SIMILAR   SOLIDS  169 


RATIOS   RELATING   TO  ANY   TWO   SIMILAR   SOLIDS 

396,  Corresponding  Cross  Sections.  Now  consider  any  two 
similar  figures  whatever  so  placed  as  to  have  a  center  of  simili- 
tude 0.  We  have  seen  that  if  points  A,  B  and  A',  B'  are  corre- 
sponding points  of  the  two  figures,  then  the  ratio  of  the 
corresponding  linear  dimensions  AB  and  A'B'  is  equal  to  the 
ratio  of  similitude  m  :  n  of  the  two  figures. 

Also  if  A,  B,  C,  D  and  A',  B',  C',  D'  are  corresponding 
points,  then  A  ABC  and  A'B'C',  and  the  tetrahedrons  ABCD 
and  A'B'C'D'  are  similar,  and  we  have 

area  ABC  =  m*  and    vol.  ABCD       m3 


area^'^B'C"      w2  vol.  A'B'C'D'      n* 

The  points  A,  B,  C  and  A1,  B',  C'  determine  two  planes,  each 
of  which  intercepts  a  certain  plane  figure  in  the  solid  figure 
to  which  the  points  belong.  These  two  plane  figures  we  call 
corresponding  cross  sections. 

We  assume  without  full  argument : 

397.  THEOKEM  X.  (1)  The  ratio  of  the  areas  of 
any  pair  of  corresponding  cross  sections  or  any  pair 
of  corresponding  surfaces  of  similar  figures  is  equal 
to  the  square  of  their  ratio  of  similitude,  and 

(2)  The  ratio  of  the  volumes  of  any  two  similar  fig- 
ures is  equal  to  the  cube  of  their  ratio  of  similitude. 

The  fact  that  the  ratio  of  the  areas  of  corresponding  surfaces 
of  similar  solids  is  equal  to  the  square  of  their  ratio  of  simili- 
tude, while  the  ratio  of  their  volumes  equals  the  cube  of  this 
ratio  is  one  of  the  most  important  and  far-reaching  conclusions 
of  geometry. 

Thus  the  ratio  of  the  weights  of  two  similar  shells  used  in 
gunnery,  or  the  ratio  of  the  weights  of  two  men  of  similar 
build,  may  be  found  when  their  ratio  of  similitude  is  known. 


170  SOLID   GEOMETRY:    APPENDIX  I 

PROBLEMS  AND  APPLICATIONS 

1.  If  it  is  known  that  a  steel  wire  of  radius  r  will  carry  a 
certain  weight  w,  how  great  a  weight  will  a  wire  of  the  same 
material  carry  if  its  radius  is  2  r  ? 

Suggestion.    The  tensile  strengths  of  wires  are  in  the  same  ratio  as 
their  cross-section  areas. 

2.  Find  the  ratio  of  the  diameters  of  two  wires  of  the  same 
material  if  one  carries  twice  the  load  of  the  other ;  three  times 
the  load. 

3.  In  a  laboratory  experiment  a  heavy  iron  ball  is  sus- 
pended by  a  steel  wire.     In  suspending  another  ball  of  twice 
the  diameter  a  wire  of  twice  the  radius  of  the  first  one  is  used. 
Is  this  perfectly  safe  if  it  is  known  that  the  first  wire  will  just 
safely  carry  the  ball  suspended  from  it  ?     Discuss  fully. 

4.  In  two  schoolrooms  of  the  same  shape  (similar  figures) 
but  of  different  size,  the  same  proportion  of  the  floor  space  is 
occupied  by  desks.     Which  contains  the  larger  amount  of  air 
for  each  pupil  ? 

5.  If  the  shells  used  in  guns  are  similar  in  shape,  find  the 
ratio  of  the  total  surface  areas  of  an  eight-inch  and  a  twelve- 
inch  shell. 

6.  Find  the  ratio  of  the  weights  of  the  shells  in  the  preced- 
ing problem,  weights  being  in  the  same  ratio  as  the  volumes. 

7.  If  a  man  5  ft.  9  in.  tall  weighs  165  lb.,  what  should  be 
the  weight  of  a  man  6  ft.  1  in.  tall,  if  they  are  similar  in  shape  ? 

8.  What  is  the  diameter  of  a  gun  which  fires  a  shell  weigh- 
ing twice  as  much  as  a  shell  fired  from  an  eight-inch  gun,  sup- 
posing the  shells  to  be  similar  bodies  ? 

9.  Supposing  two  trees  to  be  similar  in  shape,  what  is  the 
diameter  of  a  tree  whose  volume  is  three  times  that  of  one 
whose  diameter  is  2  feet  ?    What  is  the  diameter  if  the  volume 
is  five  times  that  of  the  given  tree  ?    What  if  it  is  n  times 
that  of  the  given  tree  ? 


SIMILAR   SOLIDS  171 


10.  Assuming  that  the  weights  of  schoolboys  vary  as  the 
cubes  of   their  heights,  construct  a   graph  representing  the 
relation  between  their  heights  and  weights,  if  a  boy  5  feet  9 
inches  tall  weighs  130  pounds. 

Suggestion.  If  w  represents  the  number  of  pounds  in  weight  and  h 
the  number  of  feet  in  height,  w  =  kh3.  From  w  =  130,  when  h  =  5f ,  we 
have  k  =  .684.  For  the  purpose  of  the  graph,  k  =  .7  is  accurate  enough. 

11.  From  the  graph  constructed  in  the  preceding  example 
find  the  weight  of  a  boy  5  feet  tall ;  one  5  feet  4  inches ;  one 
5  feet  6  inches.     Compare  with  the  weights  of  boys  in  your 
class. 

12.  If  a  man  6  feet  tall  weighs  185  pounds,  construct  a  graph 
representing  the  weights  of  men  of  similar  build  and  of  various 
heights. 

13.  If  steamships  are  of  the  same  shape,  their  tonnages  vary 
as  the  cubes  of  their  lengths.    The  Mauretania  is  790  feet  long, 
with  a  net  tonnage  of  32,500.     Construct  a  graph  representing 
the  tonnage  of  steamships  of  the  same  shape,  and  of  various 
lengths. 

Other  ships  which  at  one  time  or  another  have  held  ocean 
records  are  :  the  (former)  Deutschland,  length  686  ft.  and  ton- 
nage 16,500 ;  the  (former)  Kaiser  Wilhelm  der  Grosse,  length 
648  ft.  and  tonnage  14,300 ;  the  Lucania,  length  625  ft.  and 
tonnage  13,000  (nearly)  ;  and  the  Etruria,  length  520  ft.  and 
tonnage  8000.  By  means  of  this  graph  decide  whether  or  not 
these  boats  have  greater  or  less  tonnage  than  the  Mauretania 
as  compared  with  their  lengths. 

14.  Raindrops  as  they  start  to  fall  are  extremely  small.     In 
the  course  of  their  descent  a  great  many  are  united  to  form 
larger  and  larger  drops.     If  1000  such  drops  unite  into  one, 
what  is  the  ratio  of  the  surface  of  the  large  drop  to  the  sum 
of  the  surfaces  of  the  small  drops  ? 


172  SOLID  GEOMETRY:    APPENDIX  I 

15.  The  strength  of  a  muscle  varies  as  its  cross-section  area, 
which  in  turn  varies  as  the  square  of  the  height  or  length  of 
an  animal,  while  the  weight  of  the  animal  varies  as  the  cube  of 
its  height  or  length.     Use  these  facts  to  explain  the  greater 
agility  of  small  animals.      For  example,  compare  the  rabbit 
and  the  elephant. 

16.  Assuming  the  velocities  the  same,  the  amounts  of  water 
flowing  through  pipes  vary  directly  as  their  cross-section  areas. 
How  many  pipes,  each  4  in.  in  diameter,  will  carry  as  much 
water  as  one  pipe  72  in.  in  diameter  ? 

17.  What  must  be  the  diameter  of  a  cylindrical   conduit 
which  will  carry  enough  water  to  supply  ten  circular  intakes 
each  8  feet  in  diameter  ? 

18.  A  water  reservoir,  including  its  feed  pipes,  is  replaced 
by  another,  each  of  whose  linear  dimensions  is  twice  the  cor- 
responding dimensions  of  the  first.      If  the   velocity  of  the 
water  in  the  feed  pipes  of  the  new  system  is  the  same  as  that 
in  the  old,  will  it  take  more  or  less  time  to  fill  the  new  reser- 
voir than  it  did  the  old  ?     What  is  the  ratio  of  the  new  time 
to  the  old  ? 

19.  If  two  engine  plants  are  exactly  similar  in  shape,  but 
each  linear  dimension  in  one  is  three  times  the  corresponding 
dimension  of  the  other,  and  if  the  steam  in  the  feed  pipes  flows 
with  the  same  velocity  in  both,  compare  the  speeds  of  the 
engines. 

20.  If  two  men,  one  5  ft.  6  in.  and  the  other  6  ft.  2  in.  in 
height,  are  similar  in  structure  in  every  respect,  how  much 
faster  must  the  blood  flow  in  the  larger  person  in  order  that 
the  body  tissues  of  both  shall  be  supplied  equally  well  ? 

Suggestion.  Note  that  the  amount  of  tissue  to  be  supplied  varies  as 
the  cube  of  the  height,  while  the  cross-section  area  of  the  arteries  varies 
as  the  square  of  the  height. 


APPLICATIONS   OF  PROJECTION 


173 


APPENDIX   II:    PROJECTION   OF   LINE-SEGMENTS 

398,  Length  of  Projection.     The  projection  of  a  line-segment 
on  a  plane  was  shown  in  §  121  to  be  another  line-segment. 
The  length  of  this  projection  will  now  be  computed  in  terms  of 
the  given  line-segment. 

399,  Cosine  of  Projection  Angle.     The  acute  angle  between  a 
line-segment  and  a  given  line  on  which  it  is  projected  is  called 
the  projection  angle. 

If  I  is  the  length  of  a  line-segment  and  p  the  length  of  its 
projection,  then  the  ratio  p  :  Zis  called  the  cosine  of  the  projec- 
tion angle. 

E.g.,  in  Fig.  1,  -  =  cosine  /.  BAE. 


E 


F 


C 
FIG.  1 


P 


D 


b 
FIG.  2 


400,  Sine,  Cosine,  Tangent.  In  any  right  triangle  ABO 
(Fig.  2),  either  acute  angle,  as  /.  A,  is  the  projection  angle 
between  the  hypotenuse  and  the  side  adjacent  to  the  angle. 

Hence  the  cosine  of  an  acute  angle  of  a  right  triangle  is  the 
ratio  of  the  adjacent  side  to  the  hypotenuse. 

Likewise  we  define  the  sine  of  an  acute  angle  of  a  right  tri- 
angle as  the  ratio  of  the  opposite  side  to  the  hypotenuse,  and  the 
tangent  of  an  acute  angle  of  a  right  triangle  as  the  ratio  of  the 
opposite  side  to  the  adjacent  side. 

Using  the  common  abbreviations,  sin,  cos,  and  tan,  we  have 
in  Fig.  2, 

-,  cosA  =  -,  tan^  =  - 
c  c  b 


sn  -    =  - 


174 


SOLID   GEOMETRY:    APPENDIX  II 


EXERCISES  ON  SINES,   COSINES,   TANGENTS 

The  sine,  cosine,  and  tangent  are  of  great  importance  in 
many  computations.  By  careful  measurement  (and  in  other 
ways)  their  values  may  be  computed  for  any  acute  angle,  and 
a  table  formed,  like  that  on  page  175. 

jE.gr.,  if  Z.A  =  35°  (measured  with  a  protractor),  and  if  we  measure 
J.C,  AB,  and  BC,  and  then  compute  the  ratios 

-,  -,  and  -,  we  shall  find  the  values  of  sin  35°, 
c    c  b 

cos  35°,  tan  35°. 

With  an  ordinary  ruler  it  will  not  usually  be 
possible  to  make  these  measurements  with  suffi- 
cient accuracy  to  obtain  more  than  one  decimal     •*  6       C 
place. 

EXERCISES 

1.  Using  a  protractor,  construct  angles  of  10°,  30°,  50°,  70°, 
and  by  measurement  determine  the  sine,  cosine,  and  tangent 
of  each. 

2.  Prove  that  the  cosine  of  any 
given  angle  is  the  same,  no  matter 
what  point  is  taken   in   either   side 
from  which  to  let  fall  the  perpen- 
dicular to  the  other  side.     Prove  the  same  for  the  tangent. 

3.  Show  that  if  the  hypotenuse  be  taken  one  decimeter  in 
length,  then  the   length   of  the  side  adjacent,  measured  in 
decimeters,  is  the  cosine  of  the  angle,  and  the  length  of  the 
side  opposite  is  the  sine  of  the  angle. 

4.  Show  that  if  the  side  adjacent  be  taken  one  decimeter  in 
length,  the  length  of  the  side  opposite,  measured  in  decimeters, 
is  the  tangent  of  the  angle. 

5.  Without  any  direct  measurement,  show  how  to  compute 
the  three  ratios  for  each  of  the  angles  30°,  45°,  60°. 

Suggestion.  Make  use  of  the  fact  that  if  one  acute  angle  in  a  right 
triangle  is  30°,  the  side  opposite  it  is  one  half  the  hypotenuse. 


APPLICATIONS   OF  PROJECTION 


175 


TABLE  OF  SINES,   COSINES,   AND  TANGENTS 


Angle 

Sin 

Cos 

Tan 

Angle 

Sin 

Cos 

Tan 

0° 

0 

1.000 

0 

46° 

.719 

.695 

1.04 

1° 

.017 

1.000 

.017 

47° 

.731 

.682 

1.07 

2° 

.035 

.999 

.035 

48° 

.743 

.669 

1.11 

3° 

.052 

.999 

.052 

49° 

.755 

.656 

1.15 

4° 

.070 

.998 

.070 

50° 

.766 

.643 

1.19 

5° 

.087 

.996 

.087 

51° 

.777 

.629 

1.23 

6° 

.105 

.995 

.105 

52° 

.788 

.616 

1.28 

7° 

.122 

.993 

.123 

53° 

.799 

.602 

1.33 

8° 

.139 

.990 

.141 

54° 

.809 

.588 

1.38 

9° 

.156 

.988 

.158 

55° 

.819 

.574 

1.43 

10° 

.174 

.985 

.176 

56° 

.829 

.559 

1.48 

11° 

.191 

.982 

.194 

57° 

.839 

.545 

1.54 

12° 

.208 

.978 

.213 

58° 

.848 

.530 

1.60 

13° 

.225 

.974 

.231 

59° 

.857 

.515 

1.66 

14° 

.242 

.970 

.249 

60° 

.866 

.500 

1.73 

15° 

.259 

.966 

.268 

61° 

.875 

.485 

1.80 

16° 

.276 

.961 

.287 

62° 

.883 

.469 

1.88 

17° 

.292 

.956 

.306 

63° 

.891 

.454 

1.96 

18° 

.309 

.951 

.325 

64° 

.899 

.438 

2.05 

19° 

.326 

.946 

.344 

65° 

.906 

.423 

2.14 

20° 

.342 

.940 

.364 

66° 

.914 

.407 

2.25 

21° 

.358 

.934 

.384 

67° 

.921 

.391 

2.36 

22° 

.375 

.927 

.404 

68° 

.927 

.375 

2.48 

23° 

.391 

.921 

.424 

69° 

.934 

.358 

2.61 

24° 

.407 

.914 

.445 

70° 

.940 

.342 

2.75 

25° 

.423 

.906 

.466 

71° 

.946 

.326 

2.90 

26° 

.438 

.899 

.488 

72° 

.951 

.309 

3.08 

27° 

.454 

.891 

.510 

73° 

.956 

.292 

3.27 

28° 

.469 

.883 

.532 

74° 

.961 

.276 

3.49 

29° 

.485 

.875 

.554 

75° 

.966 

.259 

3.73 

30° 

.500 

.866 

.577 

76° 

.970 

.242 

4.01 

31° 

.515 

.857 

.601 

77° 

.974 

.225 

4.33 

32° 

.530 

.848 

.625 

78° 

.978 

.208 

4.70 

33° 

.545 

.839 

.649 

79° 

.982 

.191 

5.14 

34° 

.559 

.829 

.675 

80° 

.985 

.174 

5.67 

35° 

.574 

.819 

.700 

81° 

.988 

.156 

6.31 

36° 

.588 

.809 

.727 

82° 

.990 

.139 

7.12 

37° 

.602 

.799 

.754 

83° 

.993 

.122 

8.14 

38° 

.616 

.788 

.781 

84° 

.995 

.105 

9.61 

39° 

.629 

.777 

.810 

85° 

.996 

.087 

11.43 

40° 

.643 

.766 

.839 

86° 

.998 

.070 

14.30 

41° 

.656 

.755 

.869 

87° 

.999 

.052 

19.08 

42° 

.669 

.743 

.900 

88° 

.999 

.035 

28.64 

43° 

.682 

.731 

.933 

89° 

1.000 

.017 

57.29 

44° 

.695 

.719 

.966 

90° 

1.000 

0 

45° 

.707 

.707 

1.00 

176  SOLID   GEOMETRY:    APPENDIX  II 

LENGTH   OF  PROJECTION  OF  A  LINE-SEGMENT 

401,  THEOREM  I.  The  length  of  the  projection  of  a 
line-segment  upon  a  given  line  is  equal  to  the  length  of 
the  line-segment  multiplied  ~by  the  cosine  of  the  projec- 
tion angle. 

TD 

Given  the  projection  p  of  the 
line-segment  /  on  the   line  CZ>, 


with  the  projection  angle  A.  *^      \    P 

To  prove  that  p  =  I  cos  A. 
Proof :  By  definition  we  have 


=  cos  A.    Hence,  p  =  I  cos  A. 


EXERCISES 

1.  Find  the  cosines  of  the  angles  35°  30',  54°  15',  15°  45  '= 

Suggestion.  The  cosine  of  35°  30'  lies  between  cos  35°  and  cos  36°. 
We  assume  that  it  lies  halfway  between  these  numbers.  This  assump- 
tion, while  not  quite  correct,  is  very  nearly  so  for  small  differences  of 
angles,  as  in  this  case,  where  the  total  difference  is  only  one  degree. 
From  the  table  cos  35°  =  .819,  cos  36°  =  .809.  The  number  midway 
between  these  is  .814,  which  we  take  as  the  cosine  of  35°  30'. 

This  process  is  called  interpolation.  A  similar  process  is  used  for 
sines  and  tangents. 

2.  Find  the  tangents  of  the  angles  25°  20',  47°  45',  63°  40'. 

3.  Find  the  angle  whose  tangent  is  1.74. 

Solution.  From  the  table  we  have  tan  60°  =  1.73  and  tan  61°  =  1.80. 
Hence,  the  required  angle  must  lie  between  60°  and  61°.  Moreover,  the 
number  1.74  is  one  seventh  the  way  from  1.73  to  1.80.  Hence,  we 
assume  the  angle  to  lie  one  seventh  the  way  from  60°  to  61°,  which  gives 
60°  +  \  x  1°  =  60°  +  9'  nearly.  The  required  angle  is  60°  9'. 

4.  Find  the  angles  whose  sines  are  .276  ;  .674  ;  .437. 

5.  Find  the  angles  whose  cosines  are  .940  ;  .094  ;  .435. 

6.  Find  the  angles  whose  tangents  are  .781  ;  1.41  ;  3.64. 


APPLICATIONS  OF  PROJECTION 


177 


7.  At  what   angle   with   the   horizontal 
must  the  base  of  a  right  circular  cylinder 
be  tilted  to  make  it  just  topple  over  if  its 
diameter  is  6  ft.  and  its  altitude  8  ft.  ? 

Suggestion.  The  center  of  gravity  is  at  the 
middle  point  C  of  the  axis  of  the  cylinder.  The 
base  must  be  tilted  so  that  the  line  AC  becomes 
vertical.  The  required  angle  is  Z.ACB. 

8.  The  Leaning  Tower  of  Pisa  is  179  feet 
high  and  31  feet  in  diameter.     It  now  leans 
so  that  a  plumb  line  from  the  top  on  the 
lower  side  reaches  the  ground  14  feet  from 
the  base.     At  what  angle  is  its   side   now 
inclined  from  the  vertical  ?     At  what  angle 
would   its    side    have  to   incline   from   the 
vertical  before  it  would  topple  over  ? 

9.  A  four-inch  hole  is  cut  in  a  board,  and  a  ball  8  in.  in 
diameter  is  made  to  rest  on  it.     At 

what  angle  must  the  board  be  held 
so  that  the  ball  will  just  roll  out 
of  the  hole? 

Suggestion.     The  board  must  be  held 
so   that  the  line   OA  becomes  vertical. 

10.  Using   a  ball  8  inches  in  diameter,  what  must  be  the 
radius  of  the  hole  in  the  board  of  the  preceding  problem  so 
that  the  ball  shall  just  roll  out  when 

the  board  is  inclined  at  an  angle   of 
45°  to  the  horizontal  ? 

11.  If    the    figure   ABCD-H  is   a 
cube,    find    each     of     the     following 
angles :  Z.  EC  A,  Z  AEG. 

Check  by  using  the  fact  that  the  sum 
of  the  angles  of  a  triangle  is  180°. 


E 


H 


178  SOLID   GEOMETRY:    APPENDIX  II 

ALTITUDE  OF   OBLIQUE  PRISM  OR   CYLINDER 

402,  THEOREM  II.  The  altitude  of  an  oblique  prism 
or  cylinder  is  equal  to  an  element  multiplied  by  the 
cosine  of  the  angle  between  the  plane  of  the  base  and 
that  of  a  right  section. 


Given  an  oblique  prism  or  cylinder  with  base  b  and  right  section 
c,  and  let  BE  be  a  perpendicular  between  the  planes  of  the  bases. 

Consider  the  plane  determined  by  BE  and  the  element  AB. 

This  plane  is  _L  to  the  plane  of  b  and  also  to  the  plane  of  c. 
(Why  ?) 

Hence,  it  is  J_  to  the  line  of  intersection  of  the  planes  of 
b  and  c.  (Why  ?) 

Let  this  plane  cut  the  planes  of  b  and  c  in  GD  and  FD 
respectively. 

Then  Z  D  is  the  measure  of  the  dihedral  angle  between  the 
planes  of  b  and  c.  (Why  ?) 

To  prove  that  BE  =  AB  •  cos  D. 

Proof :  We  have  BE  =  AB  •  cos  Z  ABE.                    Why  ? 

But  /.D  =  ^ABE.                                  Why? 

Hence,  BE  =  AB  -  cos  D. 

403,  COROLLARY.  The  dihedral  angle  between  the  planes  of 
the  base  and  a  right  section  of  an  oblique  cylinder  or  prism  is 
equal  to  the  angle  between  an  element  and  the  altitude. 


APPLICATIONS  OF  PROJECTION 


179 


EXERCISES 

1.  Given  a  line-segment  10  inches  long.     Find  the  length 
of  its  projection  on  a  plane  if  the  projection  angle  is  20°.     If 
the  angle  is  30°,  45°,  60°,  90°,  0°. 

2.  A  kite,  string  forms  an  angle  of  40°  with  the  ground. 
The  distance  from  the  end  of  the  string  to  a  point  directly 
beneath  the  kite  is  200  ft.     Find  the  length  of  the  string  and 
the  perpendicular  height  of  the  kite. 

3.  The  altitude  of  an  oblique  prism  is  15  inches.     Find 
the  length  of  an  element  if  it  makes  an  angle  of  45°  with  the 
perpendicular  between  the  bases. 

4.  A  right  section  of  a  cylinder  makes  an  angle  of  20°  with 
the  plane  of  the  lower  base.     Find   the   ratio    between   the 
altitude  and  an  element. 

5.  Prove  that  by  joining  the 
middle  points  of  six  edges  of  a 
cube,  as  shown  in  the  figure,  a 
regular  hexagon  is  formed. 

6.  Prove  that  in  the  preced- 
ing example   the   plane   of   the 
regular  hexagon,  KLMNOP,  is 
perpendicular    to    the    diagonal 
DF  of  the  cube. 

7.  How  large  a  cube  will  be 

required  from  which  to  cut  a  stopper  for  a  hexagonal  spout, 
each  of  whose  sides  is  4  inches  ? 

8.  In  the  figure  find  the  angle  KQH. 

Suggestion.     Let  a  be  a  side  of  the  cube.    Compute  ZBT,  EQ,  and 
HQ  in  terms  a.     Note  that  Z.  QKH  =  rt.  /.. 

9.  Find  the  area  of  the  projection  of  the  hexagon  KLMNOP 
on  the  face  BCGF.    Note  that  this  projection  equals  the  whole 
square  less  A  NCO  +  A  KEL.     See  §  404.     Find    the  area  of 
the  hexagon  in  terms  of  the  side  a  of  the  cube. 


180 


SOLID  GEOMETRY:    APPENDIX  II 


AREA  OP  THE  PROJECTION  OF  A  PLANE-SEGMENT 

404,  Projection  of  a  Plane-Segment.     If  from  each  point  in 
the  boundary  of  a  plane-segment  a  perpendicular  is  drawn  to 
a  given  plane,  the  locus  of  the  feet  of  these  perpendiculars  will 
bound  a  portion  of  the  plane,  which  is  called  the  projection  of 
the  plane-segment  on  the  given  plane. 

E.g.,  the  plane-segment  A'B'  C'  in  the  plane 
Nis  the  projection  of  the  plane-segment  ABC 
from  the  plane  M  upon  N. 

The  angle  of  projection  is  the  angle 
between  the  planes  M  and  N. 

405,  THEOREM  III.     The  area  of  the  projection  of  a 
plane-segment  on  a  plane  is  equal  to  the  area  of  the 
plane-segment  multiplied  %  the  cosine  of  the  projection 
angle. 


-*-A 


Proof:  Let  the  boundary  of  the  given  plane-segment  b  be 
any  convex  polygon  or  closed  curve. 

Using  a  line  perpendicular  to  the  given  plane  as  a  gener- 
ator, develop  a  prismatic  or  cylindrical  surface  of  which  b 
is  a  section.  The  given  plane  will  cut  this  surface  in  a 
right  section  whose  area  we  denote  by  c. 

Now  cut  the  surface  by  a  plane  parallel  to  6,  forming  the 
upper  base  b'  of  a  prism  or  cylinder  whose  altitude  is  7i,  edge 
e,  and  volume  V. 


APPLICATIONS  OF  PROJECTION 


181 


Then  c  is  the  projection  of  b  upon  the  given  plane,  and 
Z  1  =  Z  2  is  the  projection  angle. 
We  are  to  show  that  c  =  b  cos  Z  1. 

We  know  that  F=  ce  =  6fc.  Why  ? 

But  /f  =  e  cos  Z  2. 

Hence,  ce  =  be  cos  Z  2. 

That  is,  c  =  6  cos  Z  2  =  6  cos  Z  1. 

Note.  The  foregoing  theorem  may  be  proved  directly  in  case  the 
plane-segment  is  a  rectangle  with  one  side  par- 
allel to  the  line  of  intersection  of  the  two  planes. 
In  the  figure  let  S  be  the  given  rectangle  and  S' 
its  projection,  with  AB  II  to  the  line  of  intersec- 
tion of  the  planes  in  which  S  and  Sf  lie,  and  Z 1 
the  angle  between  them. 

Then  S  =  AB-BC 

and  S'=A'B''B'C'. 

But  AB  =  A'B' 

and  BE=B'C'.                             Why? 

But  BE=BC-cosZl                  Why? 

and       S'  =  A'B'.  B' C'  =  AB  •  BCcosZl. 

That  is,  8'  =  S  cos  Z  1. 

In  the  case  of  any  plane-segment,  rectangles  may  be  inscribed  in  it 
in  this  position  and  their  number  increased  indefinitely,  so  that  their  sum 
will  approach  more  and  more  nearly  to  the  area  of  the  plane-segment, 
and  in  this  way  it  may  be  shown  to  any  desired  degree  of  approximation, 
that  the  projection  of  a  plane-segment  equals  the  given  plane-segment 
multiplied  by  the  cosine  of  the  projection  angle. 

406,  Ellipse.  An  important  special  case  of  the  theorem  of 
§  405  is  the  area  of  the  figure  obtained  by  projecting  a  circle 
upon  a  plane  not  parallel  to  its  own  plane,  nor  at  right  angles 
to  it.  This  figure  is  called  an  ellipse.  On  page  182  the  prin- 
ciple developed  above  is  used  to  find  the  area  of  the  ellipse. 


182 


SOLID   GEOMETRY:    APPENDIX  II 


THE   AREA  OF   AN   ELLIPSE 

407,  Projection  of  a  Circle.  In  the  figure  two  planes,  M  and 
M' ,  meet  in  a  line  PQ.  The  circle  0  in  M  has  a  diameter 
u4B  II  PQ  and  a  diameter  <7Z>  J_  PQ. 

In  projecting  the  whole  figure  upon  the  plane  M1  the 
diameter  AB  projects  into  its  equal  A'B',  while  CD  projects 
into  C'D'  so  that  C'D'  =  CD  cos  /.  1. 


By  theorem  §  405  the  area  of  the  ellipse  A'C'B'D'  equals  the 
area  of  the  circle  ACBD  multiplied  by  cos  Z  1. 

Hence,  -nr2  -  cos  Z  1  is  the  area  of  the  ellipse. 

But  r  cos  Z  1  =  O'C'  and  r  =  O'B'.  (§  401) 

Hence,  the  area  of  the  ellipse  is  ?r  •  O'C'  x  O'B'. 

The  segments  A'B'  and  C'D'  are  called  respectively  the 
major  and  minor  axes  of  the  ellipse,  and  O'B'  and  O'C1  the 
semimajor  and  the  semiminor  axes.  These  latter  are  usually 
denoted  by  a  and  b. 

Hence,  the  area  of  the  ellipse  is  irab. 

Note  that  when  a  and  b  are  equal,  the  ellipse  becomes  a  circle,  and 
this  formula  reduces  to  ?ra2  as  it  should. 


APPLICATIONS   OF  PROJECTION  183 

PROBLEMS   AND   APPLICATIONS 

1.  Given  a  right  circular  cylinder  the  radius  of  whose  base 
is  6  inches.     Find  the  area  of  an  oblique  cross  section  inclined 
at  an  angle  of  45°  to  the  plane  of  the  base. 

2.  Given  an  oblique  circular  cylinder  the  radius  of  whose 
right  section  is  10  inches.     Find  the  area  of  the  base  if  it  is 
inclined  at  an  angle  of  60°  to  the  right  section. 

3.  If  an  oblique  circular  cylinder  has  an  altitude  h,  an  ele- 
ment e,  radius  of  right  section  r,  and  Z.  A  the  inclination  of 
the  base  to  the  right  section,  express  the  volume  in  two  ways 
and  show  that  these  are  equivalent. 

4.  A   six-inch   stovepipe  has  a  45°  elbow 
angle,  that  is,  it  turns  at  right  angles.     (The 
angle  CAB  is  called  the  elbow  angle.)     Find 
the  area  of  the  cross  section  of  AB.     Likewise 
if  it  has  a  60°  elbow  angle. 

5.  At  what  angle  must  the  damper  in  a 

circular  stovepipe  be  turned  in  order  to  obstruct  just  half  the 
right  cross  sectional  area  of  the  pipe  ? 

Suggestion.     The  damper  must  be  turned  so  that  the  projection  of 
the  damper  upon  a  right  cross  section  is  equal  to  half  that  cross  section. 

6.  The  comparatively  low  temperature  of  the  earth's  surface 
near  the  pole,  even  in  summer,  when  the  sun  does  not  set  for 
months,  is  due  partly  to  the  obliqueness  with  which  the  sun's 
rays  strike  the  earth.     That  is,  a  given  amount  of  sunlight  is 
spread  over  a  larger  area  than  in  lower  latitudes. 

Thus,  if  in  the  figure  D'C  is  a  horizontal  line,  and  D'D  the  direction 
of  the  sun's  rays,  then  a  beam  of  light 
whose  right  cross  section  is  ABCD 
is  spread  over  the  rectangle  A' BCD'. 
In  other  words,  a  patch  of  ground 
A' BCD'  receives  only  as  much  sun- 
light as  a  patch  the  size  of  ABCD 
receives  when  the  sun's  rays  strike  it 
vertically.  ABCD  =  A' BCD'  cos  Z.  1 . 


184 


SOLID   GEOMETRY:    APPENDIX  II 


Hence,  each  unit  of  area  in  A' BCD'  receives    cosZl    times   as    much 
light  as  a  unit  in  ABCD. 

Hence,  to  compare  the  heat-producing  powers  of  sunlight  in  any  lati- 
tude with  that  at  the  place  where  the  sun's  rays  fall  vertically,  we  need 
to  know  how  the  projection  angle,  Z 1,  is  related  to  the  difference  in  lati- 
tude of  the  two  places. 

7.  If  Z  1  =  30°,  compare  the  amount  of  heat  received  by  a 
unit  of  area  in  ABCD  and  A' BCD'. 

8.  What  must  Z  1  be  in  order  that  a  unit  of  area  in  A' BCD' 
shall  receive  only  -J-  as  much  light  as  a  unit  in  ABCD  ? 

9.  The  figure  represents  a  cross  section  of  the  earth  with  an 
indication  of  the  direction  of  the  rays  of  light  as  they  strike 
it  at  the  summer  solstice  when  they  are  vertical   at  A,  the 
tropic  of  Cancer.     B  represents  the  latitude  of  Chicago,  C  the 
polar  circle,  and  P  the  north  pole.     The  angles  PDE,  CGF, 
BKH  represent   the   projection   angle,    Z  1,  for   the  various 
latitudes.     Prove : 

Z  PDE  =  Z  POTT, 
Z  CGF  =  Z  CO K, 
Z  BKH  =  Z  BOK. 


That  is,  Z 1  for  each  place 
is  the  latitude  of  that  place 
minus  the  latitude  of  the 
place  where  the  sun's  rays 
are  vertical. 

10.  Find  the  relative  amount  of  sunlight  received  by  a  unit 
of  area  at  the  tropic  of  Cancer  and  at  the  north  pole  at  the 
time  of  the  summer  solstice. 

11.  Find  the  ratio  between  the  amount  of  light  received  by 
a  unit  of  the  earth's  area  at   Chicago  and  at   the   tropic   of 
Cancer  at  the  time  of  the  summer  solstice. 

12.  Find  the  same  ratio  for  the  equator  and  Chicago  at  the 
winter  solstice  when  the  sun  is  vertical  at  latitude  23^°  south. 


THEORY  OF  LIMITS  185 

APPENDIX   III:  VARIABLES.     LIMITS 

408,  Variables  and  Functions.     It  is  often  useful  to  think  of 
a  geometric  figure  as  varying  in  size  and  shape. 

E.g.,  if  a  rectangle  has  a  fixed  base,  say  10  inches  long,  but  an  altitude 
which  varies  from  3  inches  to  5  inches,  then  the  area  varies  from  3  •  10  = 
30  to  5  •  10  =  50  square  inches. 

We  may  even  think  of  the  altitude  as  starting  at  zero  inches  and 
increasing,  in  which  case  the  area  starts  at  zero  and  increases  continuously. 

The  altitude  which  we  think  of  as  varying  at  our  pleasure  is 
called  the  independent  variable,  while  the  area,  being  dependent 
upon  the  altitude,  is  called  the  dependent  variable. 

The  dependent  variable  is  sometimes  called  a  function  of  the 
independent  variable,  meaning  that  the  two  are  connected  by 
a  definite  relation  such  that,  for  any  definite  value  of  the  inde- 
pendent variable,  the  dependent  variable  also  has  a  definite 
value. 

Thus,  in  the  formula  for  the  area  of-  a  rectangle,  a  =  bh,  if  b  is  fixed 
and  h  varies,  then  a  is  a  function  of  h,  since  for  every  value  of  h  there  is 
determined  a  definite  value  of  a. 

409,  Illustrations  of  Limits.     If  a  regular  polygon  is  inscribed 
in  a  circle  of  fixed  radius,  and  if  the  number  of  sides  of  the 
polygon  be  continually  increased,  for  instance  by  repeatedly 
doubling  the  number,  then  the  apothem,  perimeter,  and  area 
are  all  variables  depending  upon  the  number  of  sides.     That  is, 
each  of  these  is  a  function  of  the  number  of  sides. 

Now  the  greater  the  number  of  sides  the  more  nearly  does 
the  apothem  equal  the  radius  in  length.  Indeed,  it  is  evident 
that  the  difference  between  the  apothem  and  the  radius  will 
ultimately  become  less  than  any  fixed  number,  however  small. 
Hence  we  say  that  the  apothem  approaches  the  radius  as  a  limit 
as  the  number  of  sides  increases  indefinitely. 


186  SOLID   GEOMETRY:    APPENDIX  III 

Similarly  the  perimeters  of  the  polygons  may  be  made  as 
nearly  equal  to  the  circumference  as  we  please  by  making  the 
number  of  sides  sufficiently  great. 

Hence  we  may  define  the  circumference  of  a  circle  as  the 
limit  of  the  perimeter  of  a  regular  inscribed  polygon  as  the 
number  of  sides  increases  indefinitely. 

The  circumference  of  a  circle  may  also  be  defined  as  the 
limit  of  the  perimeter  of  a  circumscribed  polygon  as  the  num- 
ber of  sides  is  increased  indefinitely. 

Likewise  we  may  define  the  area  of  a  circle  as  the  limit  of 
the  area  of  the  inscribed  or  the  circumscribed  polygon  as  the 
number  of  sides  is  increased  indefinitely. 

The  notion  of  a  limit  may  be  used  to  define  the  length  of  a 
line-segment  which  is  incommensurable  with  a  given  unit  seg- 
ment. 

Thus,  the  diagonal  d  of  a  square  whose  side  is  unity  is  d  =  \/2.  Hence 
d  may  be  defined  as  the  limit  of  the  variable  line-segment  whose  succes- 
sive lengths  are  1,  1.4,  1.41,  1.414, .... 

In  like  manner,  the  length  of  any  line-segment,  whether 
commensurable  or  incommensurable  with  the  unit  segment, 
may  be  defined  in  terms  of  a  limit. 

Thus,  if  a  variable  segment  is  increased  by  successively  adding  to  [it 
one  half  the  length  previously  added,  then  the  segment  will  approach  a 
limit.  If  the  initial  length  is  1,  and  if  the  successive  additions  are  £,  £,  \, 
^,  -5^,  etc.,  then  the  successive  lengths  are  1,  1£,  If,  1£,  1}|,  lf£,  etc. 
Evidently  this  segment  approaches  the  limit  2. 

Hence  2  may  be  defined  as  the  limit  of  the  variable  segment  whose 
successive  lengths  are  1,  1£,  If,  1|,  etc.,  as  the  number  of  successive  addi- 
tions is  increased  indefinitely. 

The  idea  of  a  functional  relation  between  variables  and  the 
idea  of  a  limit,  as  illustrated  above,  are  two  of  the  most  im- 
portant concepts  in  all  mathematics. 


THEORY  OF  LIMITS  187 

DEFINITION  OF  A  LIMIT 

410,  Constant.     A   quantity   which  remains  fixed  in  value 
throughout  a  discussion  is  called  a  constant. 

E.g.,  the  base  of  the  rectangle  mentioned  in  §  408  is  a  constant. 

411,  Variable.     A  quantity  which  continuously  changes  in 
value,  or  which  takes  on  a  succession  of  different  values,  is 
called  a  variable. 

E.g.,  the  altitude  and  the  area  mentioned  in  §  408  are  variables. 

412,  Limit  of  a  Variable.     If  a  variable  may  be  made  to 
approach  a  certain  constant  quantity  in  such  a  way  that  the 
difference  between  the  constant  and  the  variable  becomes  and 
remains  less  than  any  assignable  value,  however  small,  then 
the  constant  is  called  the  limit  of  the  variable. 

E.g.,  the  fixed  circumference  of  the  circle  is  the  limit  of  the  variable 
perimeters  of  the  polygons  mentioned  in  §  409. 

SIGHT  WORK 

1.  Find  the  limit  of  a  variable  line-segment  whose  initial  length  is  6 
inches,  and  which  varies  by  successive  increments  each  equal  to  one  hah5 
the  preceding,  the  first  increment  being  2  inches. 

2.  Construct  a  right  triangle  whose  sides  are  1  and  2.     By  approximat- 
ing a  square  root,  find  five  successive  lengths  of  a  segment  which  ap- 
proaches the  length  of  the  hypotenuse  as  a  limit. 

3.  If  one  tangent  to  a  circle  is  fixed  and  another  is  made  to  move  so 
that  their  intersection  point  approaches  the  circle,,  what  is  the  limiting 
position  of  the  moving  tangent  ?     What  is  the  limit  of  the  measure  of  the 
angle  formed  by  these  tangents  ? 

4.  The  arc  AB  of  74°  is  the  greater  of  the  two  arcs  intercepted  be- 
tween two  secants  meeting  at  C  outside  the  circle.     The  points  A  and  B 
remain  fixed  while  C  moves  up  to  the  circle.     What  angle  is  the  limit  of 
the  variable  angle  formed  by  the  varying  secants  ?    What  is  the  limit 
of  the  measure  of  this  angle  ? 

5.  If  in  the  preceding  the  secants  meet  within  the  circle,  what  is  the 
limit  of  their  angle  and  also  of  the  measure  of  this  angle  as  the  intersection 
point  moves  up  to  the  center  of  the  circle  ? 


188  SOLID   GEOMETRY:    APPENDIX  III 

THE  INCOMMENSURABLE  CASES 

413,  The  Incommensurable  Cases.     We  have  seen  that  there 
are  segments  which  are  incommensurable;  that  is,  which  have 
no  common  unit  of  measure,  —  for  instance,  the  side  and  the 
diagonal  of  a  square. 

For  practical  purposes  the  lengths  of  such  segments  are  ap- 
proximated to  any  desired  degree  of  accuracy,  and  their  ratios 
are  understood  to  be  the  ratios  of  these  approximate  numerical 
measures. 

But  for  theoretical  purposes  it  is  important  to  consider  these 
incommensurable  cases  further ;  just  as  in  algebra  we  not  only 
approximate  such  roots  as  V2,  V3,  V5,  etc.,  but  we  also  deal 
with  these  surds  as  exact  numbers. 

Instances  of  this  kind  occur  in  such  operations  as 

(V3  +  \/2)(V3 -V2)=3-2  =  l. 

While  the  length  of  the  diagonal  of  a  unit  square  cannot  be 
expressed  as  an  integer  or  as  a  rational  fraction,  that  is,  as  the 
quotient  of  two  integers,  we  nevertheless  think  of  such  a  seg- 
ment as  having  a  definite,  length,  or  what  is  the  same  thing,  a 
definite  ratio  with  the  unit  segment  forming  the  side  of  the 
square. 

We  now  fix  our  attention  on  the  incommensurable  ratios 
themselves,  and  the  method  of  determining  them,  rather  than 
on  the  process  of  approach  and  the  practical  computation  based 
on  it. 

414,  Representation  by  Irrational  Numbers.     In  general,  the 
ratio  between  any  two  incommensurable  geometric  magnitudes 
of  the  same  kind  may  be  represented  by  what  is  called  an 
irrational  number;  that  is,  a  number  which  is  neither  an  integer 
nor  a  quotient  of  two  integers. 

Examples  of  irrational  numbers  are  the  surds,  such  as  V2 
-\/5,  etc.,  and  the  number  rr. 


THEORY  OF  LIMITS  189 


NUMBERS  DEFINED  BY  SEQUENCES 

415,  Numbers  Defined  by  Sequences.  The  following  is  a 
method  for  determining  any  number,  whether  rational  or 
irrational.  For  simplicity  it  is  applied  first  to  the  integer  1. 

.Throughout  this  discussion  the  expressions  "  point  on  a  line  " 
and  "  number  "  will  be  used  interchangeably. 

On  a  straight  line  mark  a  certain  point  0  (zero),  and  one  unit  to  the 

right  of  it  mark  another  point  1.  Also  , , , , r 

lay  off  points  such  that  their  distances  °  ill* 

from  0  are,  £,  |,  },  i|,  .... 

If  this  sequence  of  points  is  carried  ever  so  far,  it  will  never  reach  the 
point  1.  If,  however,  we  select  a  point  to  the  left  of  1,  no  matter  how 
near  to  it,  we  may  always  go  far  enough  along  this  sequence  to  reach 
points  between  this  point  and  1.  That  is,  1  is  the  limit  of  the  terms  of 
this  sequence.  See  §  412. 

The  point  1  has  two  definite  relations  to  this  sequence : 

(a)  Every  point  of  the  sequence  is  to  the  left  o/l. 

(b)  For  any  fixed  point  to  the  left  of  1  there  are  points  of  the 
sequence  between  it  and  1. 

We  see  that  1  is  the  only  point  on  the  whole  line  such  that 
both  (a)  and  (b)  are  true  of  it.  For  every  point  to  the  right 
of  1  (a)  is  true,  but  (b)  is  not.  For  every  point  to  the  left  of 
1  (b)  is  true,  but  (a)  is  not. 

It  follows  therefore  that,  while  the  points  of  the  sequence 
merely  approach  the  point  1  as  a  limit,  the  sequence,  taken  as 
a  whole,  serves  to  distinguish  that  point  from  all  other  points  on 
the  line. 

In  the  above  sequence,  the  terms  continually  increase.  The  number  1 
may  be  determined  equally  well  by  a  decreasing  sequence.  For  instance, 
the  sequence  1^,  1£,  1£,  Ij1^,  •••  has  the  following  properties  which  distin- 
guish the  point  1  from  all  other  points  on  the  line. 

(a)  Every  point  of  the  sequence  is  to  the  right  of  1. 

(b)  For  any  fixed  point  to  the  right  of  1  there  are  points  of  the 
sequence  between  it  and  1. 


190  SOLID   GEOMETRY:    APPENDIX  III 

LIMIT  OF  AN  INFINITE  SEQUENCE 

416,  Infinite  Sequences.     An  endless  sequence  of  either  kind 
just  described  is  called  an  infinite  sequence. 

Not  every  infinite  sequence  serves  to  single  out  a  definite 
point  in  the  manner  shown  above.  Thus  the  sequence  1,  2,  3> 
4,  •••  fails  to  do  so,  because  its  terms  grow  large  beyond  all 
bound.  Such  sequences  are  said  to  be  unbounded,  while  the 
sequences  -J,  f,  J,  •«  and  1^,  1J,  1|,  •••  are  bounded. 

Again,  the  sequence  1,  2,  1,  2,  1,  2,  —  fails  to  single  out  a 
definite  point.  This  sequence  is  said  to  be  oscillating,  since  its 
terms  increase,  then  decrease,  then  increase,  etc.,  while  -J,  J 
-J,  •••  and  1-J-,  11,  1-J-,  •••  are  non-oscillating. 

417,  Limit  of  a  Sequence.     The  number  1  is  said  to  be  the 
least  upper  bound  of  the  sequence  -J-,  f ,  J-,  — .     That  is,  1  is  the 
smallest  number  beyond  which  the  sequence  does  not  go.     1  is 
also  said  to  be  the  limit  of  the  sequence.     See  §  412. 

Similarly,  1  is  the  greatest  lower  bound  or  the  limit  of  the 
sequence  1J,  1-J-,  1-J-,  »• ;  that  is,  the  greatest  number  such  that 
the  sequence  contains  no  number  less  than  it. 

In  the  manner  described  above  any  integer  may  be  deter- 
mined by  an  increasing  or  a  decreasing  infinite  sequence. 

E.g.,  the  number  2  is  the  limit  of  the  increasing  sequence  1, 1£,  If,  1|, 
1H»  •"»  or  °f  the  decreasing  sequence  2£,  2^,  2£,  2T^,  •••. 

Two  different  sequences,  both  increasing  or  both  decreasing, 
may  also  define  the  same  number. 

E.g.,  each  of  the  increasing  sequences  £,  f,  |,  ^|,  •••  and  f,  f ,  f,  f,  ••• 
determines  the  number  1. 

We  notice,  however,  that  no  matter  what  definite  number  we  select  in 
either  of  these  sequences,  there  is  a  number  in  the  other  greater  than  it ; 
that  is,  neither  sequence  contains  a  number  greater  than  every  number  in 
the  other.  See  §  420. 

It  is  apparent  that  any  number  may  be  defined  as  the  limit 
of  a  sequence,  and  we  shall  assume  the  converse,  namely,  that 
any  bounded  increasing  or  decreasing  sequence  has  a  limit. 


THEORY  OF  LIMITS  191 

FUNDAMENTAL  PRINCIPLES  OP  LIMITS 

418,  AXIOM.     We  now  assume  that :  Every  bounded  increas- 
ing sequence  has  a  least  upper  bound,  and  every  bounded  decreas- 
ing sequence  has  a  greatest  lower  bound. 

This  axiom  may  also  be  stated  : 

Every  bounded  increasing  or  decreasing  sequence  has  a  limit. 

This  axiom  simply  means  that  every  such  sequence  singles 
out  a  definite  number,  rational  or  irrational,  in  the  manner 
discussed  on  page  190. 

Thus,  if  we  attempt  to  approximate  the  square  root  of  2,  we  obtain  a 
sequence,  1,  1.4,  1.41,  1.414,  1.4142,  ••-,  having  for  its  limit  a  definite 
number  represented  by  V2,  which  corresponds  to  the  length  of  the  diago- 
nal of  a  square  whose  side  is  unity. 

Again,  if  we  attempt  to  approximate  the  value  of  ?r,  we  obtain  a 
sequence  3,  3.1,  3.14,  3.141,  3.1415,  •••,  having  for  its  limit  the  definite 
number  represented  by  ?r,  which  is  the  ratio  of  the  circumference  to  the 
diameter  of  a  circle. 

419,  THEOREM  I.     Two  bounded  sequences  have  the 
same  limit  if  they  are  equal  term  ~by  term. 

Proof  :  If  the  sequences  are  cti,  a2,  a3,  •••  and  bl9  b2,  b3,  •»,  and 
if  a1  =  &i,  a2  =  b2,  a3  =  &3,  •••,  then  they  are  one  and  the  same 
sequence  and  hence  have  the  same  limit  by  §  418. 

Historical  Note.  The  rigorous  treatment  of  limits  has  been  developed 
during  the  last  fifty  years.  The  foundation  on  which  it  rests  is  the  theory 
of  irrational  numbers,  which  itself  was  placed  on  a  firm  basis  in  the  early 
seventies. 

The  Greeks  dealt  with  the  incommensurable  cases  in  an  interesting 
manner.  Thus,  to  prove  that  two  incommensurable  ratios  are  equal  they 
showed  that  neither  can  be  less  than  the  other.  However,  they  failed  to 
make  explicit  definitions  of  incommensurables  and  they  did  not  explicitly 
state  the  fundamental  axiom  of  §  418  or  its  equivalent.  Their  treatment 
would  be  complete  and  rigorous  if  supplemented  by  the  proper  defini- 
tions and  axioms. 


192  SOLID   GEOMETRY :    APPENDIX  III 

FUNDAMENTAL   PRINCIPLES  OF  LIMITS 

420,  THEOREM  II.     Two  increasing  bounded  sequences 
have  the  same  limit  if  neither  sequence  contains  a  num- 
ber greater  than  every  number  of  the  other. 

Let  «!,  a2,  a3,  •••  and  61?  62,  6S,  •••  denote  two  infinite  sequences 
with  limits  a  and  b,  such  that  no  term  of  the  first  sequence  is 
greater  than  every  term  of  the  second,  and  no  term  of  the 
second  is  greater  than  every  term  of  the  first. 

To  prove  that  a  =  b. 

Proof :  Suppose  that  a  is  not  equal  to  b  but  is  less  than  b. 
Then  there  must  be  numbers  of  the  sequence  b1}  b2,  o3)  —  greater 
than  a,  since  by  §  415  there  are  numbers  b1}  b2,  b3)  greater  than 
any  fixed  number  whatever  which  is  less  than  b.  This  contra- 
dicts the  hypothesis  of  the  theorem,  and  hence  a  cannot  be 
less  than  b.  In  like  manner  we  show  that  b  is  not  less  than  a. 
Hence  a  =  b.  See  §  20. 

421,  THEOREM   III.     Two    decreasing    bounded    se- 
quences have  the  same  limit  if  neither  sequence  contains 
a  number  less  than  every  number  of  the  other. 

Proof :     The  proof  is  exactly  similar  to  that  of  Theorem  II. 

EXERCISES 

1.  Show  that  the  sequences  1,  f ,  £,  —  and  f,  f ,  ^,  —  have  the 
same  limits. 

2.  Show  that  the  sequences  -J-,  £,  -$., ...  and  £,  |,  |,  ».  have  the 
same  limits. 

The  applications  of  the  theory  of  limits  to  geometry  consist 
chiefly  in  showing  that  two  numbers  are  equal  because  they  are 
the  limits  of  the  same  sequence,  or  of  sequences  having  the 
property  stated  in  the  theorem  of  §  420  or  of  §  421. 


THEORY  OF  LIMITS  193 

APPLICATION  OF  LIMITS  TO  PLANE  GEOMETRY 

422,  PROBLEM.  On  a  given  segment  AB  to  lay  off  a 
sequence  of  points  Bly  B2)  B^--,  of  which  B  is  the  limit, 
such  that  each  of  the  segments  ABly  AB2,  ABS, ...  is 
commensurable  with  a  given  segment  CD. 


nig  w»i 


•fc 


Solution  ?  Using  mi,  an  exact  divisor  of  CD,  as  a  unit  of  measure, 
lay  off  on  AB  a  segment  AB^,  an  exact  multiple  of  mi,  such  that  the 
remainder  B\B  is  less  than  mi.  Then  CD  and  ABi  are  commensurable. 

Using  a  unit  m2,  likewise  a  divisor  of  CD,  and  less  than  B^B,  lay  off 
ABZ  such  that  B2B  is  less  than  m2.  Then  AB2  >  ABi,  and  CD  and  AB2 
are  commensurable. 

Continuing  in  this  manner,  using  as  units  of  measure  segments  ma, 
m/4,  •«•  each  an  exact  divisor  of  CD  and  each  less  than  -B2.B,  B3B,  •  •• 
respectively,  we  obtain  a  sequence  of  segments  AB±,  AB^  AB$,  •••,  each 
greater  than  the  preceding  and  each  commensurable  with  CD. 

If  the  units  mi,  m2,  ma,  -••  are  so  selected  that  they  approach  zero  as  a 
limit,  it  follows  that  B  is  the  limit  point  of  the  sequence  JBi,  B%,  B%,  ••>. 

If  a  different  sequence  of  divisors  of  CD,  as  m^,  m2',  ra3',  •••,  is  used, 
we  obtain  a  sequence  of  points  BI',  -B2',  2V,  •••,  likewise  satisfying  the 
conditions  of  the  problem. 

Any  two  such  sequences  of  points  Z?b  B2,  U3,  •••  and  IV,  -82',  -83',  ••• 
determined  as  above,  are  both  increasing  and  each  is  such  that  no  point 
of  it  is  to  the  right  of  every  point  of  the  other. 

Hence,  by  §  420  any  two  such  sequences  have  the  same  limit  point  B. 

423,   Limit  of  a  Sequence  of  Segments.     If  Bl}  B2,  B3, ...  is  a 

sequence  of  points  on  the  segment  AB  having  the  limit  B,  then 
the  segment  AB  is  said  to  be  the  limit  of  the  sequence  of  segments 
AB1}  AB2,  AB3, .-. 


194  SOLID   GEOMETET:    APPENDIX  III 


INCOMMENSURABLE  RATIOS 

424,  Ratio  of  Two  Incommensurable  Segments.     The  ratio  of 
two  incommensurable  segments  has  not  been  explicitly  denned, 
but  this  may  now  be  done  in  terms  of  the  limit  of  a  sequence. 

Consider  two  incommensurable  segments  AB  and  CD.  Let 
a-i,  a2,  «s,  •••  be  the  lengths  of  segments  each  commensurable 
with  CD,  forming  a  sequence  whose  limit  is  the  segment  AB, 
and  let  b  be  the  length  of  the  segment  CD. 

Then  ^i,  ^.  ^,  ...  is  an  increasing  bounded  sequence  having 
b     b     b 

a  limit  which  we  call  R. 

If  a/,  a2',  a3',  •••  are  the  lengths  of  segments  forming  any 

CL  '     a  '    a  ' 
other  sequence  whose  limit  is  AB,  the  sequence  -^     -^  ,  -£•  ,  —  is 

b       b      b 
another  increasing  bounded  sequence  with  limit  R'. 

By  §  420  we  now  know  that  R  =  R'. 

This  number  R  is  denned  as  the  ratio  of  the  incommensurable 
segments  AB  and  CD. 

425.  Ratio  of  Two  Incommensurable  Arcs.     The  considerations 
of  §§  422,  423,  424  apply  to  two  arcs  of  equal  circles. 


Thus,  in  the  figure,  the  ratio  of  the  two  incommensurable  arcs  AB 


and  CD  is  the  limit  of  the  sequence        ±,         3,  ,  —  ,  and  the  ratio 

CD       CD      CD 

of  the  incommensurable  angles  AOB  and  COD  is  the  limit  of  the  sequence 
ZAOBi     ZAOB2     ZAOBs 
Z  COD  '    Z.  COD  '    Z  COD  '  "*' 


THEORY  OF  LIMITS  195 


APPLICATIONS   OF  LIMITS   TO  PLANE  GEOMETRY 

426,  THEOREM  IV.  A  line  parallel  to  the  base  of  a 
triangle,  and  meeting  the  other  two  sides,  divides  them 
in  the  same  ratio. 

Given  the  A  ABC  with  DE  I!  EC  and 
cutting  AB  and  AC. 

To  prove 


Proof  :  Consider  the  case  when  AD 
and  AB  are  incommensurable. 

Let  Dl}  DZ,  Dz,  •»  be  a  sequence 
of  points  on  AB  whose  limit  is  D. 
Through  these  points  draw  parallels     ° 
to  BC,  meeting  AC  in  E},  E2,  Es,  —  . 

Then  E  is  the  limit  of  the  sequence  of  points  El}  Ez,  E3,  •». 

Tor  suppose  it  is  not,  and  that  there  is  a  point  K  on  AE 
such  that  there  is  no  point  of  EI,  E2,  Ez,  •••  between  ^Tand  E. 
Draw  a  line  parallel  to  BC  through  K,  meeting  AB  in  H. 

'  But  there  are  points  of  the  sequence  Z>b  D2,  D3,  between  H 
and  D,  and  hence  there  must  be  points  of  the  sequence  EI,  E2, 
EZ,  •••  between  K  and  E,  which  shows  that  E  is  the  limit  of 
the  sequence  EI,  E»,  Es,  •••. 


Then  the  sequence  (1)  ~1,  ,  ,  ...  has  a  limit  R 

which,  by  definition,  is  the  ratio  of  the  segments  AD  and  AB. 

§424 


Similarly,  the  sequence  (2)          ,  ,  ,  ..,  has  a  limit 

AC>      AC      AC 

R'  which  is  the  ratio  of  AE  to  AC. 


Hence,  the  two  sequences,  (1)  and  (2),  are  identical  and  have 
the  same  limit  (§  419).     Hence  R  =  Rr.  Q.  E.  D. 


196 


SOLID   GEOMETRY:    APPENDIX  III 


APPLICATIONS    TO   PLANE    GEOMETRY 

427,  THEOREM  V.  In  the  same  circle  or  in  equal 
circles  the  ratio  of  two  central  angles  is  the  same  as  the 
ratio  of  their  intercepted  arcs. 


Proof:  In  case  the  arcs  are  commensurable  the  proof  is 
obvious.  For,  in  that  case,  the  common  measure  of  the  arcs 
divides  each  into  a  number  of  equal  parts,  say  m  and  n  respec- 
tively, and  these  arcs  subtend  the  same  number  of  equal  angles 
at  the  center  (§  244,  Plane  Geometry.)  Hence  the  ratio  of  the 
arcs  is  m  :  n  and  the  ratio  of  the  angles  is  also  m  :  n. 

If  the  arcs  AB  and  CD  are  not  commensurable,  let  ABi, 
AB2,  ABS,  —  be  a  sequence  of  arcs  whose  limit  is  AB,  each 
arc  being  commensurable  with  the  arc  CD. 


Then  the  sequence 


jj*, 


,  ~  has  a  limit  B  = 


0.    .,    ,    ^  Z.AOBl     ^AOB2 

Similarly  the  sequence  —^,  -^^ 


CD 

,  •••  has 


a  limit  R'  = 


But 


Z<70Z> 

AB1==ZAOBl    AB«^. 
CD      Z  COD  '    CD      Z  COD ' 


as  shown  above  for  the  commensurable  case. 

Hence,  these  two  sequences  are  identical  and  have  the  same 
limit  (§  419).  Therefore  it  follows  that  R  =  R'.  Q.  E.  D. 

Note.  —  The  student  should  note  the  successive  steps  in  the  proofs  of 
§§  426,  427  :  (a)  Definitions  of  the  incommensurable  ratios  by  means 
of  infinite  sequences ;  (6)  Proof  that  these  sequences  are  identical ; 
(c)  Conclusion  that  the  limiting  ratios  are  equal. 


THEORY  OF  LIMITS 


197 


428,  Products  of  Irrational  Numbers.     If  al}  a2,  a3,   •••  is  a 
sequence  with  limit  a,  then  ka1}  fca2,  kaS)  •••  is  a  sequence  whose 
limit  is  defined  as  ka  where  k  is  any  number. 

If  ab  02,  a3,  —  and  61?  52,  &3>  •••  are  two  sequences  with  limits 
a  and  6  respectively,  then  o^,  a2b2,  a363,  —  is  a  sequence  whose 
limit  is  defined  as  ab. 

Similarly  if  o1?  a2,  a8,  »•,  &b  62>  &s?  •••  an(i  ci>  C2>  cs>  •••  are 
sequences  with  limits  a,  6,  c,  then  abc  is  defined  as  the  limit 
of  the  sequence  a^Ci,  a2&2c2>  ^b3c3)  •». 

For  a  complete  treatment  it  would  be  necessary  to  show  that 
these  definitions  of  multiplication  of  irrational  numbers  are 
consistent  with  the  rest  of  arithmetic  and  also  that  these  new 
sequences  are  such  as  to  determine  definite  limits.  This  is 
possible  but  is  beyond  the  scope  of  this  book. 

429,  Area  of  a  Rectangle.     If  the  sides  of  a  rectangle  are  in- 
commensurable with  the  unit  segment,  we  define  its  area  as 
follows : 

Let  a1?  a2,  03,  •••  be  a  sequence  of  rational  numbers  whose 
limit  is  the  altitude  a ;  and  let  61?  62,  63,  —be  a  sequence  whose 
limit  is  the  base  b. 

Then  the  area  of  the  rectangle  is  the  limit  of  the  sequence  o^, 


a3 
a, 

a< 

a 

b 

But  by  definition  the  limit  of  a^,  a262,  a363,  —  is  the  prod- 
uct ab.     Hence  we  have  the  theorem : 

430,  THEOREM  VI.      The  area  of  a  rectangle  is  the 
product  of  its  base  and  altitude. 


198  SOLID   GEOMETRY:    APPENDIX  III 

APPLICATIONS    TO   PLANE  GEOMETRY 

431,  Incommensurable  Ratios  Related  to  Two  Circles.     In  a 
circle  inscribe  a  sequence  Pl9  P2,  P3,  •••  of  regular  polygons, 
each,  having  twice  the  number  of  sides  of  the  one  preceding  it. 

Let  the  perimeters  of  these  polygons  be  plt  p2)  p^,  •••  and 
their  areas  Ai9  A2,  A3,  —.  Then  the 
length  c  of  the  circle  is  denned  to  be 
the  limit  of  the  sequence  pl9  p2,  p3)  ••« 
and  its  area  A  is  denned  to  be  the 
limit  of  AI,  A2,  A3)  •». 

The  sequence  of  polygons  thus  in- 
scribed is  called  an  approximating 
sequence  of  polygons. 

That  these  sequences  are  increasing 
and  bounded  is  obvious  from  the  figure. 

432,  THEOREM  VII.     The  lengths  of  two  circles  are 
in  the  same  ratio  as  their  radii,  and  their  areas  are  in 
the  same  ratio  as  the  squares  of  their  radii. 

Proof :  Let  the  radii  of  the  two  circles  be  r  and  /.  Denote 
the  ratio  r1 :  r  by  k.  Then  r'  =  kr. 

Inscribe  in  one  circle  an  approximating  sequence  of  polygons 
with  perimeters  pl9  p2)  p3)  «««  and  areas  Ai9  A2)  As,  •••.  In  the 
other  circle  inscribe  a  sequence  of  similar  polygons.  By 
§§  476,  477  (Plane  Geometry),  the  perimeters  of  the  latter  are 
kpi,  kpz)  kps)  "-  and  their  areas  are  kzAl9  k2A2)  k2A3,  •». 

By  §  428,  if  the  limit  of  ply  p2,  p3,  •••  is  c  and  the  limit  of 
Al9  A2,  As,  —  is  Aj  then  the  limits  of  kph  kp2)  kps,  ••-,  and 
k2A1}  k2A2,  k2A3,  —  are  kc  and  k2A,  respectively. 

That  is,  the  ratio  of  the  lengths  of  the  circles  is  —  =  k  =  — 

c  r 

and  the  ratio  of  their  areas  is  —=  =  k2  =  —  • 


THEORY  OF  LIMITS 


199 


APPLICATIONS   OF   LIMITS   TO   SOLID   GEOMETRY 

433,  Incommensurable  Solids.     The  areas  of  curved  surfaces, 
the  volumes  of  cones,  cylinders,  and  spheres,  and  even  of  rec- 
tangular solids  involve  incommensurable  quantities.     We  have 
already  treated  these  cases  informally  and  have  devised  means 
for  computing  them  approximately. 

We  now  give  a  formal  treatment  for  logical  completeness. 

434,  Rectangular   Parallelepiped.      If   the   three    concurrent 
edges  a,  6,  c  of  a  rectangular  parallelepiped  are  incommen- 
surable  with   the   unit   segment,  the 

volume  inclosed  is  denned  as  follows : 

Let  ttj,  «2>  as>  •••  be  a  sequence  of 
rational  numbers  whose  limit  is  the 
side  a.  Let  bit  b2,  63,  —  and  Cj,  c2,  c3, 
».  be  similar  sequences  whose  limits 
are  respectively  the  dimensions  b  and  c. 

Then  the  volume  is  the  limit  of  the 
sequence  a^Ci,  a2b2c2,  a363c3, 

But  the  limit  of  this  sequence  is  by 

definition  the  product  of  the  limits  of  the  three  sequences 
«i>  «2>  03,  •-,  &i,  &2>  &3>  —,  Ci,  c2,  c3,  ..-,  or  abc. 

Hence,  we  have  the  theorem : 

435,  THEOREM   VIII.     The  volume  of  a  rectangular 
parallelepiped   is  equal   to   the  product  of  its    three 
concurrent  edges. 

EXERCISES 

1.  Give  in  decimals  four  terms  of  a  sequence  which  approxi- 
mates   the    volume   of   a    rectangular    parallelepiped  whose 
dimensions  are  2,  4,  V5. 

2.  G-ive  in  decimals  four  terms  of  a   sequence  which  ap- 
proximates the  circumference  of  a  circle  whose  radius  is  10 
in. ;  also  of  a  sequence  which  approximates  the  area. 


200  SOLID   GEOMETRY:    APPENDIX  III 

APPLICATION   OF   LIMITS   TO  THE  PYRAMID 

436,  Triangular  Pyramid.     In  a  triangular  pyramid  inscribe 
a  set  of  prisms  in  a  manner  similar  to  that  shown  in  the  figure 
of  §  243. 

Call  FI  the  sum  of  the  volumes  of  these  inscribed  prisms. 

Using  as  altitudes  one  half  the  altitudes  of  the  first  set  of 
prisms,  inscribe  a  second  set  the  sum  of  whose  volumes  is  V2. 
Continuing  in  this  manner,  we  obtain  a  sequence  of  sets  of 
prisms  with  volumes  FI,  F2,  F3,  — .  The  limit  F  of  this  se- 
quence we  define  as  the  volume  of  the  pyramid. 

Circumscribed  prisms  may  also  be  used  for  defining  the  volume  of  a 
pyramid,  in  which  case  we  get  a  decreasing  sequence  of  volumes 
#i,  #2,  #3,  •••»  with  limit  U.  That  U  =  V  follows  from  the  fact  that 
the  sum  of  the  volumes  of  the  circumscribed  prisms  exceeds  that  of  the 
inscribed  prisms  by  the  volume  of  the  lowest  circumscribed  prism,  and 
this  may  be  made  as  small  as  we  please. 

437,  THEOREM  IX.     Two  pyramids  with  the,  same 
altitudes  and  equivalent  bases  have  equal  volumes. 

Proof :  Call  a  sequence  of  inscribed  prisms  in  one  pyramid 
FI,  F2,  F3,  •••,  and  in  the  other  FI',  F2',  F3',  •  •-.  Since  corre- 
sponding sets  of  these  prisms  have  equal  volumes  (§  200),  that 
is,  Fj=  F/,  F2=F2',  F3  =  F3'  •••,  the  theorem  follows  from 
§419. 

438,  Convex  Closed  Curves.      We  assume  that  in  a  convex 
closed  curve  (§  202)  it  is  possible  to  inscribe  a  sequence  of 
polygons  PJ,  P.,,  P3,  — ,  having  perimeters  p^  p2,  p3,  —  and 
areas  Ai}  A2,  A3,  —  with  limits  p  and  A  respectively,  and  to 
circumscribe  a  sequence  of  polygons  P/,  P2',  P3',  •»,  having 
perimeters  pj,  p2,  p3',  —  and  areas  AJ,  A2'}  A3',  —  with  limits 
p'  and  A'  respectively,  such  that  p  =  pf  and  A  =  A'. 

These  limits  p  and  A  we  now  define  as  the  perimeter  and  the 
area  respectively  of  the  curve. 


THEORY  OF  LIMITS  201 

APPLICATION   OF   LIMITS    TO   THE  CYLINDER 

439,  The  Cylinder.     Given  any  cylinder  with  a  convex  right 
cross-section  and  an  element  e.     In  this  cross-section  inscribe 
a  sequence  PI,  P2,  P^  '"  °f  polygons,  as  in  §  438,  with  perime- 
ters jp1?  p2,  Psi  '"  an(i  areas  AI,  A2,  A3,  «««,  thus  defining  the 
perimeter  p  and  the  area  A  of  the  cross-section. 

Consider  a  set  of  prisms  inscribed  in  this  cylinder,  of  which 
PI,  P2,  P3,  •••  are  right  cross-sections. 

Then  the  areas  and  the  volumes  of  these  prisms  are  respec- 
tively P&,  p2e,  p3e,  -  and  A&,  A2e,  A3e,  •». 

The  lateral  area  and  the  volume  of  the  cylinder  are  now  defined 
as  the  limits  of  these  sequences.  But  by  §  428  these  limits  are 
equal  respectively  to  pe  and  Ae. 

Hence,  we  have  the  theorem : 

440.  THEOREM  X.     The  lateral  area  of  a  cylinder  is 
the  product  of  an  element,  and   the  perimeter   of  a 
right  section  and  its  volume  is  the  product  of  an  element 
and  the  area  of  a  right  section. 

EXERCISES 

1.  Prove  as  above  that  the  volume  of  any  convex  cylinder 
is  equal  to  the  product  of  its  altitude  and  the  area  of  its  base. 

2.  Prove  that  the  lateral  area  of  a  right  circular  cone  is 
equal  to  half  the  product  of  the  slant  height  and  the  perimeter 
of  its  base. 

3.  Prove  that  the  volume  of  any  convex  cone  is  equal  to 
one  third  the  product  of  its  altitude  and  the  area  of  its  base. 

Suggestion.  The  treatment  required  in  these  exercises  is  a  very  close 
paraphrase  of  the  definitions  and  proof  given  in  §  439.  Observe  that  we 
cannot  begin  to  make  a  proof  until  we  have  defined  the  subject  matter 
of  the  theorem.  That  is,  we  must  first  define  the  areas  and  volumes  in 
question. 


202  SOLID   GEOMETRY:    APPENDIX  III 


APPLICATION   OF   LIMITS   TO   THE   SPHERE 

441,   The  Sphere.     Through  two  points  P  and  Q  of  a  sphere 
pass  a  great  circle  forming  a  hemisphere  with  center  0. 


A' 


Divide  (X4,  the  radius  perpendicular  to  the  plane  of  the 
great  circle,  into  the  equal  parts  00,  OB,  and  BA.  Through 
O  and  B  pass  planes  parallel  to  the  plane  of  POQ,  meeting 
the  sphere  in  points  D  and  E,  respectively. 

Construct  right  circular  cylinders  with  axes  00  and  OB 
and  radii  CD  and  BE.  Denote  by  Vl  the  sum  of  the  volumes 
of  these  cylinders. 

Now  divide  the  radius  0 A  into  six  equal  parts  and  construct 
five  cylinders  in  the  same  manner  as  above.  Let  the  sum  of 
these  volumes  be  V%. 

Continuing  in  this  manner,  each  time  dividing  OA  into 
twice  as  many  equal  parts  as  in  the  preceding,  we  obtain  a 
sequence  of  sets  of  cylinders  and  a  corresponding  sequence 
Fi,  F2,  F3,  -  of  volumes. 

We  now  define  the  volume  of  the  hemisphere  as  the  limit  of 
the  sequence  FI,  F2,  F3, 

442,  Construct  a  right  circular  cylinder  with  its  base  in  the 
plane  of  POQ  and  with  radius  and  altitude  both  equal  to  OA. 

Denote  by  F  the  figure  formed  by  the  lower  base  of  the 
cylinder,  its  lateral  surface  and  the  lateral  surface  of  the  cone 
whose  base  is  the  upper  base  of  the  cylinder  and  whose  vertex 
is  at  0'.  (See  the  right-hand  figure.) 


THEORY  OF  LIMITS  203 

Draw  segments  O'M  and  O'N.  Let  the  planes  through  C 
and  B  cut  O'A'  in  C'  and  B'  and  O'Jf  in  If  and  K. 

Now  form  the  cylinder  O'C'H  whose  axis  is  O'C'  and  whose 
radius  is  OH.  Likewise  form  the  cylinder  C'B'K. 

Let  FI'  denote  the  sum  of  the  volumes  of  O'C'D'  and  CfB'E' 
minus  the  sum  of  the  volumes  of  O'C'H  smd  OB'K. 

In  a  similar  manner,  using  the  planes  which  divide  OA  and 
hence  O'A'  into  six  equal  parts,  we  form  another  set  of  five 
cylinders,  the  sum  of  whose  volumes  minus  that  of  the  smaller 
inside  cylinders  we  denote  by  F2'. 

Continuing  in  this  manner,  we  obtain  a  sequence  of  volumes 
FI',  F2',  F3',  —  whose  limit  F'  we  define  as  the  volume  of  the 
given  figure  F. 

We  now  prove  that  Vl  =  F/,  F2  =  F2',  -. 

Denote  OA  by  r,  and  note  that  O'B'  =  B'K 

(1)  Vol.    C'B'E'  -  Vol.    C'B'K  =  irOB1  (BE*  -  WK2)  = 


(2)  Vol.   CBE  =  ',rCB'BE2  =  7rCBi(r2-'0^)t  since  BE2  = 
r*  -  01?.  .  But  OB  =  O'B'  and  CB  =  C"B'. 
Hence,  Vol.  CBE  =  TrC^r2  -  WB'2). 
Hence,  Vol.  CBE  =  Vol.  C'B'E'  -Vol.  C'B'K. 

Similarly  we  show  that 

Vol.  OCZ)  =  Vol.  O'C'D'  -Vol.  O'C'H. 

Hence,  FI  =  F/.     In  like  manner  F2  =  F2',  V3  =  Vj,  •». 

Hence,  F=  F',  since  they  are  the  limits  of  the  same 
sequences. 

But  the  volume  of  the  cylinder  O'A'M  is  Trr3  and  of  the  cone 
whose  volume  was  subtracted,  J  Trr3.  That  is,  the  volume  of 
.Fis  |  Trr3,  and  hence  that  of  the  hemisphere  is  f^r2. 

Hence,  we  have  the  theorem  : 

443,  THEOREM  XI.     The  volume  of  the  sphere  is 


204  SOLID   GEOMETRY:    APPENDIX  III 

Note  that  the  above  proof  consists  essentially  in  showing 
that  the  area  of  the  circle  BE  is  equal  to  that  of  the  ring 
between  the  circles  B'E'  and  B'K,  and  that  the  area  of  the 
circle  CD  is  equal  to  that  of  the  ring  between  C'D'  and  C'H 
and  so  on. 

Indeed,  this  theorem  and  also  that  of  §  437  are  special  cases 
of  what  is  known  as  Cavalieri's  Theorem. 

444,  THEOREM  XII.     If  two  solid  figures   are  re- 
garded as  resting  on  the  same  plane  b,  and  if  in  every 
plane  parallel  to  b  the  sections  of  the  two  figures  have 
equal  areas,  the  figures  have  equal  volumes. 

The  proof  of  this  general  theorem  is  more  difficult  than  any 
thus  far  given,  inasmuch  as  it  involves  sequences  which  oscillate; 
that  is,  which  are  neither  constantly  increasing  nor  constantly 
decreasing. 

445.  The  Area  of  the  Sphere.     About  a  sphere  of  radius  r 
construct  a  sequence  of  circumscribed  polyhedrons  such  that 
the  largest  face  in  each  polyhedron  becomes  as  small  as  we 
please  when  we  proceed  along  the  sequence.     Let  «b  s2>  ss,  — 
be  the  total  surfaces  of  these  polyhedrons.     This  forms  a  de- 
creasing sequence  with  limit  S  which  we  define  as  the  surface 
of  the  sphere. 

The  volumes  of  these  polyhedrons  will  be  -J-rS],  £rs2,  ^rs3,  —  . 
Then  the  volume   V  of  the  sphere  is  defined1  as  the  limit 
of  this  sequence  of  volumes. 

Hence,  by  §  428,  V=  \rS.     But  by  §  443,  F=  f  Trr3. 
Then  S  =  * 


Hence,  we  have  the  theorem  : 

446,  THEOREM  XIII.     The  area  of  the  sphere  is 

*  This  definition  can  be  shown  to  be  consistent  with  that  of  §  441. 


THEORY  OF  LIMITS  205 

EXERCISES  ON  LIMITS 

1.  In  addition  to  those  which  are  found  in  the  text  give 
other  examples  of  infinite  sequences  which  do  not  determine 
definite  numbers. 

2.  Give   two   increasing   sequences  which  determine   the 
number  3.    Show  that  the  theorem  of  §  420  applies  and  proves 
that  these  sequences  determine  the  same  number. 

3.  Give  two  decreasing  sequences  each  of  which  determines 
the  number  5.     Apply  §  421  to  show  that  these   sequences 
determine  the  same  number. 

4.  State  fully  the  relation  between  a  bounded  increasing 
sequence  and  the  number  determined  by  it.     State  also  the 
relations  between  a  bounded  decreasing  sequence  and  the  num- 
ber determined  by  it. 

5.  State  fully  what  is  meant  by  "  a  limit  of  a  sequence  " 
both  for  increasing  and  decreasing  sequences. 

6.  Given   two   incommensurable   segments   AB  and    CD. 
Lay  off  on  the  line  AB  a  decreasing  sequence  of  segments, 
each  of  which  is  commensurable  with  CD,  such  that  the  limit 
of  the  sequence  is  the  segment  AB. 

7.  If  OH  a2,  03,  »•  is  an  increasing  sequence  defining  the 
number  4,  prove  that  3aly  3  02,  3a3,  •••  defines  the  number 
3  X  4  =  12. 

8.  If  Oi,  02,  c4,  and  61?  62>  t>3,  —  are  increasing  sequences 
defining  the  numbers  3  and  5,  show  that  the  sequence  a^, 
a^bz,  «3&3,  •••  defines  the  number  15. 

9.  Describe  in  more  detail  the  meaning  of  the  ratio  of  two 
incommensurable  arcs  as  indicated  in  §  425.     Show  that  this 
ratio  is  independent  of  the  sequence  of  units  of  measurement 
used,  so  long  as  the  limit  of  this  sequence  is  zero. 

10.  Treat  the  ratio  of  two  incommensurable  angles  in  a 
manner  similar  to  the  treatment  of  arcs  in  the  preceding 
exercise. 


INDEX 


(References  are  to  sections  unless  otherwise  stated.) 


Altitude,  of  a  cone   .    .     . 

of  a  cylinder 

of  a  frustum 

of  a  prism 

of  a  pyramid 

of  a  spherical  segment 

of  a  zone 

Angle,  between  two  curves 

between  a  line  and  a  plane 

dihedral 


of  projection 119, 

polyhedral 

spherical 

trihedral 

Approach 409, 

Arc  of  a  great  circle 

Area,  of  a  cone 

of  a  curved  surface      .... 

of  a  cylinder 

of  a  prism 

of  a  pyramid 

of  a  rectangle 

of  a  sphere 368, 

of  a  spherical  polygon      .     .     . 

of  a  spherical  triangle      .     .     . 

of  a  zone 

Axioms,  10-23,   65-67,  124-128, 

244,  265,  271,  367,  370, 
Axis,  of  a  cone 

of  a  cylinder 

of  a  circle  on  a  sphere      .     .     . 


>,  of  a  cone 

of  a  cylinder 

of  a  prism  ........ 

of  a  pyramid 

of  a  spherical  sector    .... 
of  a  spherical  segment     .     .     . 
Birectangular  spherical  tri- 
angle       


254 
206 
239 
171 
233 
372 
372 
315 
118 
105 
138 
399 
136 
315 
138 
415 
295 
265 
215 
219 
171 
237 
430 
445 
363 
362 
374 
219, 
418 
255 
209 
287 


254 
205 
165 
232 
373 
372 


356 


Bound,  greatest  lower  . 


417 


least  upper 417 

Cavalieri's  theorem   ....    444 

Center,  of  a  sphere 276 

of  similitude 386 

Circle,  axis  of 287 

great 288 

poles  of 287 

small 288 

Circular,  cone 255 

cylinder 208 

Circumscribed,  cone  .  .  264,  366 

cylinder 217 

polyhedron 307,  371 

prism  .  .  .  .  .  .  218,  243, 436 

pyramid  .  .  .  ' 264 

sphere 312 

Commensurable,  angles,  arcs  427 
segments 422, 426 


Cone,  altitude  of  . 


253,  254 


base  of 254 

circular 255 

element  of 251 

lateral  surface  of  ...  254,  266 

oblique 255 

right  circular 255,  256 

spherical 373 

vertex  of 251 

volume  of 270,  271 

Conical  surface,  element  of  .  251 

generator  of 251 

nappes  of 252 

vertex  of 251 

Constant 410 

Corresponding,  cross-sections  396 
linear  dimensions  .  .  .  387,  396 
parts  of  polar  triangles  .  .  .  338 
parts  of  similar  polyhedrons  .  382 
polyhedral  angles 320 

Cosine  of  an  angle      ....    400 


207 


208 


INDEX 


(References  are  to  sections  unless  otherwise  stated.) 


Cube 173 

Curved  surface 201 

Curves,  angle  between      ...  315 

convex,  closed     ....      202,438 

Cylinder,  bases  of    ...      204,  205 

circular 208 

circumscribed 217 

element  of 206 

inscribed 218 

lateral  surface  of 219 

right 208 

volume  of 224 

Cylindrical   surface  ....  203 

element  of 203 

generator  of 203 

Degree,  spherical 357 

Dihedral  angles 105 

bisector  of 120 

equal 108 

generation  of 107 

measure  of 113 

plane  angle  of 106 

right 109 

Distance,  between  two  points 

on  a  sphere 295,  324 

polar 297 

Dodecahedron    ....      157,  158 

Edge,  of  a  dihedral  angle  ...  105 

of  a  polyhedral  angle  ....  137 

of  a  polyhedron 155 

Element,  of  a  cone 251 

of  a  cylinder 206 

Ellipse,  area  of     ....      406,407 

Equal,  areas 354 

volumes 192 

Equivalent  solids 192 

Faces,  of  a  dihedral  angle     .     .    105 
of  a  polyhedral  angle  ....    137 

of  a  polyhedron 155 

of  a  prism 171 

of  a  pyramid 232 

Figures  in  space  .     .     .     .     2,  5,  8,  9 
on  a  sphere 5 

Foot,  of  a  line 70 

of  a  perpendicular  .     .     .     .71,  118 

Frustum,  of  a  pyramid     .     .     .    ! 
of  a  cone 258 


Function 408 

Generator,  of  a  conical  surface  251 

of  a  cylindrical  surface   .     .    .  203 

of  a  prismatic  surface      .     .     .  163 

of  a  pyramidal  surface    .     .     .  229 

of  a  spherical  surface      .     .    .  366 

Geometry,  solid 3 

Great  circle,  axis  of    ....  287 

pole  of 287 

on  a  sphere 288 

Half-plane 105 

Icosahedron 157 

Incommensurables  .  188,  413,  424, 
431,  433 

Inscribed,  cone  ....  264,  366 

cylinder 218 

prism 217,243,436 

polyhedron 308 

pyramid 264 

sphere 307 

Irrational,  number 414 

ratio 424 

Isosceles  spherical  triangle    332 

Lateral,  edges  ....  165, 232 

faces 165,  232 

surfaces 165,232 

Limit,  of  segments 423 

of  a  sequence 416 

of  a  variable 409,412 

Loci  problems,  pages  3,  39,  47,  48, 
70,  89,  103,  109,  115,  152 

Lune,  angle  of 358 

Measurement,  of  surfaces,  215,  219, 

237,  265,   362,  367,   374,  404,  407, 

429,  439,  445 

of  volumes,  184,  188,  192,  216,  224, 

244,   270,   369,    375,   434,  436,  439, 

443,445 


Nappes,  of  a  conical  surface 
of  a  pyramidal  surface    .     . 


252 
230 


Octahedron 157 

Order  of  parts,  in  a  triangle    .    140 
in  polyhedral  angle     ....    141 


INDEX 


209 


(References  are  to  sections  unless  otherwise  stated.) 


Pantograph 395 

Parallel,  line  to  a  plane  ...  87 

plane  to  a  plane 86 

Parallelepiped,  rectangular  .  173 

volume  of 184 

Perpendicular,  line  to  a  plane  71 

plane  to  a  line 71 

plane  to  a  plane 110 

Plane,  determination  of  ...  68 

projections  upon  .  .  .  118,  398 

-segment 404 

Plane  angle  of  a  dihedral  angle  106 

Polar,  distances 297 

triangles 335,  338 

Poles  of  a  circle 287 

Polygons,  spherical  ....  319 

symmetrical 325 

Polyhedral  angles  ....  136 

equal 139 

edges  of 137 

faces  of 138 

symmetrical 146 

Polyhedrons 153 

added 172 

circumscribed  ....  307,  371 

classified 157 

convex 154 

edges  of 155 

equal,  equivalent 192 

faces  of 155 

inscribed 308 

models  of 159 

regular 158 

similar 382 

surface  of 156 

vertices  of 155 

Prismatic  surface  ....  162 

generator  of 163 

Prism 164 

altitude  of 171 

area  of 171,176 

bases  of 165 

circumscribed  ....  218, 243 

hexagonal 170 

inscribed 217,  243,  436 

lateral  edges  of 165 

lateral  faces  of 171 

quadrangular 170 


Prism  —  Continued 

regular 170 

right 166 

triangular 170 

truncated 172 

volume  of  ...     184,  188,  193,  198 

Problems  and  Applications, 

pages  18,  24,  27,  39,  47,  60,  68,  76, 
78,  90,  91,  99,  103,  104,  116,  123, 
131,  140,  143,  145,  148-156,  158, 
161, 170-172, 174,  176, 177, 179, 183, 
184,  192/199,  201,  205 

Projection,  of  a  circle      ...    407 

of  a  figure 118 

of  a  line-segment    .     .     .      121,  398 
of  a  plane-segment      ....    404 

Projection  angle    .    .     .      119,  399 

Pyramid 231 

altitude  of 233 

base  of 232 

frustum  of 238 

lateral  faces  of 232 

regular 234 

triangular 232 

truncated 238 

Pyramidal  surface      ....    229 

element  of 229 

nappes  of 230 

vertex  of 229 


Quadrant 


Radius,  of  a  circular  cylinder  .  208 

of  a  sphere 277 

Ratio,  incommensurable  .  .  .  424 

of  similitude 388 

Regular,  polyhedrons  .  .  158,  161 

prisms 170 

pyramids 234 

tetrahedrons 161 

Right,  cone 255 

cylinder 208 

prism 166 

pyramid 234 

section 166,207 

Section,  of  a  cone 253 

of  a  cylinder 204 


210 


INDEX 


(References  are  to  sections  unless  otherwise  stated.) 


Section  —  Continued 

of  a  prism 164 

of  a  pyramid 231 

of  a  sphere 283 

Sector,  spherical 373 

Segment,  intercepted  by  planes     88 
spherical          372 

Sequence,  approximating      422,  431 

bounded 416 

decreasing,  increasing     .     .     .    417 

infinite 416 

limit  of 417 

numbers  defined  by     ....    415 

oscillating 416 

unbounded 416 

Sight  work,  pages  3,  13,  16,  17,  26, 
31,  34,  37,  38,  43,  45,  49,  51,  55,  56, 
58,  59,  66,  67,  70,  76,  79,  80,  83,  84, 
89,  92,  93,  97,  99,  100, 101,  109,  112, 
115, 118, 119, 120, 122,  124,  129, 130, 
135,  139,  143,  166,  187 

Similar,  cones  of  revolution  .     .    380 
cylinders  of  revolution     .     .     .    378 

figures 396 

polyhedrons 382 

Similarity    .    .      pages  168,  170,  171 

Similitude,  center  of    ....    386 
ratio  of 388 

Sine  of  an  angle 400 

Slant  height,  of  frustum        239,  259 

of  pyramid 236 

of  right  cone 257 

Small  circle  on  a  sphere  ...    288 

Solids,  equal 192 

equivalent 192 

Sphere 276 

area  of 368,445 

center  of 276 

circumscribed 312 

great  circle  of 288 

inscribed 307 

points  within,  without     ...    289 

radius  of 277 

small  circle  of 288 

tangent  to 305 

volume  of 369,  371,  443 

Spherical,  angle       315 

blackboard 302 


Spherical  —  Continued 

cone 373 

degree,  minute,  second    .     .     .  357 

excess 361 

polygon 319 

sector 373 

segment 372 

surface 279 

Summaries,  pages  46,  77,   102,  147 

Surface,  conical 251 

curved 201 

cylindrical 203 

of  a  polyhedron 156 

prismatic 162 

pyramidal 229 

spherical 279 

Symmetrical,     spherical     tri- 
angles        325 

trihedral  angles 146 

Symmetry  with   respect   to  a 

point 386 

Table  of  Sines,  Cosines,  and 

Tangents 175 

Tangent,  of  an  acute  angle  .     .  400 

to  a  cone 261 

to  a  cylinder 211 

to  a  sphere 305 

Tetrahedron 157,  233 

circumscribed 309 

inscribed 312 

regular 158 

Theorems  of  Plane  Geometry 

24-63 

Theory  of  limits    .    .  Appendix  III 

Triangles,  bi-rectangular      .     .  356 

equal 354 

isosceles 332 

polar 335 

right 331 

spherical 319 

symmetrical 325 

vertical 330 

Trihedral  angles 138 

symmetrical 146 

vertical 147 

Truncated,  prism 172 

pyramid 238 


INDEX 


211 


(References  are  to  sections  unless  otherwise  stated.) 


Variables 408,411 

Vertex,  of  a  cone 251 

of  a  polyhedral  angle  ....    137 

of  a  pyramid 229 

Vertices,  of  a  polyhedron     .    .    155 

Volume,  as  a  limit  of  a  sequence  434, 

436,441 

of  a  cone 272 

of  a  cylinder 225 

of  a  prism 184,  192 


Volume  —  Continued 

of  a  sphere 369,  443 

of  spherical  cone 375 

of  spherical  sector 375 

of  spherical  segment  ....    376 


Zone,  altitude  of 372 

bases  of 372 

of  one  base     . 372 


MATHEMATICS 


A  Source  Book  of  Problems  for  Geometry 

By  MABEL  SYKES,  of  the  Bowen  High  School,   Chicago,   Illinois, 
lamo,  half  leather,  380  pages. 

THIS   book  furnishes  a  great  number  of  practical  problems 
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design  and  in  architecture. 

Plane  Trigonometry  and  Applications 

By  Professors  E.  J.  WILCZYNSKI  and  H.  E.  SLAUGHT  of  the  University 
of  Chicago.    Cloth,  276  pages. 

THIS  book  makes  a  stronger  appeal  to  the  learner  of  Trigo- 
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Logarithmic  and  Trigonometric  Tables 

By  Professors  E.  J.  WILCZYNSKI  and  H.  E.  SLAUGHT  of  the  University 
of  Chicago.     8vo,  cloth,  117  pages. 

IN  compiling  these  tables  no  pains  have  been  spared  to  make 
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Plane   Trigonometry   and   Applications:    Elementary 
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By  Professors  E.  J.  WILCZYNSKI  and  H.  E.  SLAUGHT  of  the  Univer- 
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THIS  book  is  intended  primarily  for  secondary  schools.     The 
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84 


MATHEMATICS 


Plane  and  Spherical  Trigonometry 

By  President  ELMER  A.  LYMAN,  Michigan  State  Normal  College,  and 
Professor  EDWIN  C.  GODDARD,  University  of  Michigan.  Cloth,  149 
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MANY  American  text-books  on  Trigonometry  treat  the  solu- 
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86 


MATHEMATICS 


Logarithmic  and  Other  Mathematical  Tables 

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